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#### Question 1:

Find the domain of definition of $f\left(x\right)={\mathrm{cos}}^{-1}\left({x}^{2}-4\right)$.

#### Answer:

For ${\mathrm{cos}}^{-1}\left({x}^{2}-4\right)$ to be defined
$-1\le {x}^{2}-4\le 1\phantom{\rule{0ex}{0ex}}⇒3\le {x}^{2}\le 5\phantom{\rule{0ex}{0ex}}⇒x\in \left[-\sqrt{5},-\sqrt{3}\right]\cup \left[\sqrt{3},\sqrt{5}\right]$
Hence, the domain of .

#### Question 2:

Find the domain of $f\left(x\right)=2{\mathrm{cos}}^{-1}2x+{\mathrm{sin}}^{-1}x$.

#### Answer:

For $2{\mathrm{cos}}^{-1}2x$ to be defined.

For ${\mathrm{sin}}^{-1}x$ to be defined.

Domain of
$=\left[-\frac{1}{2},\frac{1}{2}\right]$.

#### Question 3:

Find the domain of $f\left(x\right)={\mathrm{cos}}^{-1}x+\mathrm{cos}x$.

#### Answer:

For ${\mathrm{cos}}^{-1}x$ to be defined.
$-1\le x\le 1$
Now, cosx is defined for all real values.
So, domain of cosx is R.
Domain of .

#### Question 4:

â€‹Find the principal values of each of the following:

(i) ${\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
(ii) ${\mathrm{cos}}^{-1}\left(-\frac{1}{\sqrt{2}}\right)$
(iii) ${\mathrm{cos}}^{-1}\left(\mathrm{sin}\frac{4\mathrm{\pi }}{3}\right)$

(iv) ${\mathrm{cos}}^{-1}\left(\mathrm{tan}\frac{3\mathrm{\pi }}{4}\right)$

#### Answer:

(i)  Let ${\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=y$
Then,
$\mathrm{cos}y=-\frac{\sqrt{3}}{2}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]$.
Thus,
$\mathrm{cos}y=-\frac{\sqrt{3}}{2}=\mathrm{cos}\left(\frac{5\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{5\mathrm{\pi }}{6}\in \left[0,\mathrm{\pi }\right]$
Hence, the principal value of .

(ii) Let ${\mathrm{cos}}^{-1}\left(-\frac{1}{\sqrt{2}}\right)=y$
Then,
$\mathrm{cos}y=-\frac{1}{\sqrt{2}}$
We know that the range of the principal value branch is .
Thus,
$\mathrm{cos}y=-\frac{1}{\sqrt{2}}=\mathrm{cos}\left(\frac{3\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{3\mathrm{\pi }}{4}\in \left[0,\mathrm{\pi }\right]$
Hence, the principal value of .

(iii) Let ${\mathrm{cos}}^{-1}\left(\mathrm{sin}\frac{4\mathrm{\pi }}{3}\right)=y$
Then,
$\mathrm{cos}y=\mathrm{sin}\frac{4\mathrm{\pi }}{3}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]$.
Thus,
$\mathrm{cos}y=\mathrm{sin}\frac{4\mathrm{\pi }}{3}=-\frac{\sqrt{3}}{2}=\mathrm{cos}\left(\frac{5\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{5\mathrm{\pi }}{6}\in \left[0,\mathrm{\pi }\right]$
Hence, the principal value of .

(iv) Let ${\mathrm{cos}}^{-1}\left(\mathrm{tan}\frac{3\mathrm{\pi }}{4}\right)=y$
Then,
$\mathrm{cos}y=\mathrm{tan}\frac{3\mathrm{\pi }}{4}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]$.
Thus,
$\mathrm{cos}y=\mathrm{tan}\frac{3\mathrm{\pi }}{4}=-1=\mathrm{cos}\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒y=\mathrm{\pi }\in \left[0,\mathrm{\pi }\right]$
Hence, the principal value of .

#### Question 5:

For the principal values, evaluate each of the following:

(i)  ${\mathrm{cos}}^{-1}\frac{1}{2}+2{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)$
(ii)
(iii)
(iv) ${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)+{\mathrm{cos}}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

#### Answer:

(ii)

$\therefore {\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)+2{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\frac{3\mathrm{\pi }}{2}$

$\therefore {\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)+{\mathrm{cos}}^{-1}\left(\frac{\sqrt{3}}{2}\right)=-\frac{\mathrm{\pi }}{6}$

#### Question 1:

Evaluate the following:

(i)

(ii) $\mathrm{tan}\frac{1}{2}\left({\mathrm{cos}}^{-1}\frac{\sqrt{5}}{3}\right)$
(iii) $\mathrm{sin}\left(\frac{1}{2}{\mathrm{cos}}^{-1}\frac{4}{5}\right)$
(iv) $\mathrm{sin}\left(2{\mathrm{tan}}^{-1}\frac{2}{3}\right)+\mathrm{cos}\left({\mathrm{tan}}^{-1}\sqrt{3}\right)$

#### Answer:

(i)

(ii)

(iii)

(iv)
$\mathrm{sin}\left(2{\mathrm{tan}}^{-1}\frac{2}{3}\right)+\mathrm{cos}\left({\mathrm{tan}}^{-1}\sqrt{3}\right)=\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{2×\frac{2}{3}}{1+\frac{4}{9}}\right)+\mathrm{cos}\left({\mathrm{cos}}^{-1}\frac{1}{\sqrt{1+{\left(\sqrt{3}\right)}^{2}}}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{12}{13}\right)+\mathrm{cos}\left({\mathrm{cos}}^{-1}\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{12}{13}+\frac{1}{2}\phantom{\rule{0ex}{0ex}}=\frac{37}{26}$

#### Question 2:

(i) $2{\mathrm{sin}}^{-1}\frac{3}{5}={\mathrm{tan}}^{-1}\frac{24}{7}$
(ii) ${\mathrm{tan}}^{-1}\frac{1}{4}+{\mathrm{tan}}^{-1}\frac{2}{9}=\frac{1}{2}{\mathrm{cos}}^{-1}\frac{3}{5}=\frac{1}{2}{\mathrm{sin}}^{-1}\left(\frac{4}{5}\right)$
(iii)   ${\mathrm{tan}}^{-1}\frac{2}{3}=\frac{1}{2}{\mathrm{tan}}^{-1}\frac{12}{5}$

(iv)

(v)
(vi)
(vii)
(viii)
(ix)
(x)

#### Answer:

Now,

$\therefore \frac{a-b}{1+ab}=x$

#### Question 4:

Prove that
(i) ${\mathrm{tan}}^{-1}\left(\frac{1-{x}^{2}}{2x}\right)+{\mathrm{cot}}^{-1}\left(\frac{1-{\mathrm{x}}^{2}}{2\mathrm{x}}\right)=\frac{\mathrm{\pi }}{2}$

(ii) $\mathrm{sin}\left\{{\mathrm{tan}}^{-1}\frac{1-{x}^{2}}{2x}+{\mathrm{cos}}^{-1}\frac{1-{x}^{2}}{2x}\right\}=1$

(i)

(ii)

#### Answer:

Let:
$a=\mathrm{tan}z\phantom{\rule{0ex}{0ex}}b=\mathrm{tan}y$

Then,

#### Question 6:

Show that 2 tan−1x + sin−1$\frac{2x}{1+{x}^{2}}$ is constant for x ≥ 1, find that constant.

We have

#### Question 7:

Find the values of each of the following:
(i)
(ii)

#### Answer:

(i) Let  ${\mathrm{sin}}^{-1}\frac{1}{2}=y$
Then,
$\mathrm{sin}y=\frac{1}{2}$

(ii)
We have

#### Question 8:

Solve the following equations for x:

(i)

(ii)

(iii)

(iv) 2 ${\mathrm{tan}}^{-1}$ ($\mathrm{sin}$$x$$=$$\mathrm{tan}$$-$1 (2 $\mathrm{sin}$$x$), $x\ne \frac{\mathrm{\pi }}{2}$.

(v)${\mathrm{cos}}^{-1}\left(\frac{{x}^{2}-1}{{x}^{2}+1}\right)+\frac{1}{2}{\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)=\frac{2\mathrm{\pi }}{3}$

(vi)

#### Answer:

(i) We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

(ii)

(iii) We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

(iv) 2 ${\mathrm{tan}}^{-1}$ ($\mathrm{sin}$$x$$=$$\mathrm{tan}$$-$1 (2 $\mathrm{sin}$$x$), $x\ne \frac{\mathrm{\pi }}{2}$

(v)

(vi)
${\mathrm{tan}}^{-1}\left(\frac{x-2}{x-1}\right)+{\mathrm{tan}}^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{x-2}{x-1}\right)+{\mathrm{tan}}^{-1}\left(\frac{x+2}{x+1}\right)={\mathrm{tan}}^{-1}1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{x-2}{x-1}\right)={\mathrm{tan}}^{-1}1-{\mathrm{tan}}^{-1}\left(\frac{x+2}{x+1}\right)\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{x-2}{x-1}\right)={\mathrm{tan}}^{-1}\left(\frac{1-\frac{x+2}{x+1}}{1+\frac{x+2}{x+1}}\right)\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{x-2}{x-1}\right)={\mathrm{tan}}^{-1}\left(\frac{x+1-x-2}{x+1+x+2}\right)\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{x-2}{x-1}\right)={\mathrm{tan}}^{-1}\left(\frac{-1}{2x+3}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x-2}{x-1}=\frac{-1}{2x+3}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+3x-4x-6=-x+1\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}=1+6\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=7\phantom{\rule{0ex}{0ex}}⇒x=±\sqrt{\frac{7}{2}}$

#### Question 10:

Prove that:
${\mathrm{tan}}^{-1}\frac{2ab}{{a}^{2}-{b}^{2}}+{\mathrm{tan}}^{-1}\frac{2xy}{{x}^{2}-{y}^{2}}={\mathrm{tan}}^{-1}\frac{2\mathrm{\alpha \beta }}{{\mathrm{\alpha }}^{2}-{\mathrm{\beta }}^{2}},$
where α = axby and β = ay + bx.

We know

#### Question 11:

For any a, b, x, y > 0, prove that:
$\frac{2}{3}{\mathrm{tan}}^{-1}\left(\frac{3a{b}^{2}-{a}^{3}}{{b}^{3}-3{a}^{2}b}\right)+\frac{2}{3}{\mathrm{tan}}^{-1}\left(\frac{3x{y}^{2}-{x}^{3}}{{y}^{3}-3{x}^{2}y}\right)={\mathrm{tan}}^{-1}\frac{2\mathrm{\alpha \beta }}{{\mathrm{\alpha }}^{2}-{\mathrm{\beta }}^{2}}$
where α = − ax + by, β = bx + ay

Then,

#### Question 1:

Write the value of ${\mathrm{sin}}^{-1}\left(\frac{-\sqrt{3}}{2}\right)+{\mathrm{cos}}^{-1}\left(\frac{-1}{2}\right)$

#### Question 2:

Write the difference between maximum and minimum values of sin−1 x for x ∈ [− 1, 1].

#### Answer:

The maximum value of in $x\in \left[-1,1\right]$ is at 1.
So, the maximum value is

Again, the minimum value is at $-$1.
Thus, the minimum value is

So, the difference between the maximum and the minimum value is
$\frac{\mathrm{\pi }}{2}-\left(-\frac{\mathrm{\pi }}{2}\right)=\mathrm{\pi }$

#### Question 3:

If sin−1x + sin−1y + sin−1z = $\frac{3\mathrm{\pi }}{2}$, then write the value of x + y + z.

#### Question 4:

If x > 1, then write the value of sin1$\left(\frac{2x}{1+{x}^{2}}\right)$ in terms of tan−1x.

#### Question 5:

If x < 0, then write the value of cos−1$\left(\frac{1-{x}^{2}}{1+{x}^{2}}\right)$ in terms of tan−1x.

#### Answer:

Let $x=\mathrm{tan}y$
Then,

The value of x is negative.
So, let x = $-a$ where a > 0.

$-a=\mathrm{tan}y\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(-a\right)$
Now,

#### Question 6:

Write the value of tan1x + tan−1$\left(\frac{1}{x}\right)$ for x > 0.

#### Question 7:

Write the value of tan1 x + tan−1$\left(\frac{1}{x}\right)$ for x < 0.

#### Answer:

${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$
When , then both are negative.
Let x = $-$y, y>0
Then,
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}\frac{1}{x}={\mathrm{tan}}^{-1}\left(-y\right)+{\mathrm{tan}}^{-1}\left(-\frac{1}{y}\right)\phantom{\rule{0ex}{0ex}}=-\left({\mathrm{tan}}^{-1}y+{\mathrm{tan}}^{-1}\frac{1}{y}\right)\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}\left(\frac{y+\frac{1}{y}}{1-y\frac{1}{y}}\right),y>0\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}\left(\frac{{y}^{2}+1}{0}\right)\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}\left(\infty \right)\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{\pi }{2}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\pi }{2}$

#### Question 8:

What is the value of cos−1$\left(\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)+{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{2\mathrm{\pi }}{3}\right)$?

#### Question 9:

If −1 < x < 0, then write the value of ${\mathrm{sin}}^{-1}\left(\frac{2x}{1+{x}^{2}}\right)+{\mathrm{cos}}^{-1}\left(\frac{1-{x}^{2}}{1+{x}^{2}}\right)$.

#### Answer:

Let $x=-\mathrm{tan}y$
where $0
Then,

#### Question 10:

Write the value of sin (cot−1x).

#### Answer:

We know
${\mathrm{cot}}^{-1}x={\mathrm{tan}}^{-1}\frac{1}{x}$
Now, we have

Hence,

#### Question 11:

Write the value of .

#### Answer:

We have

${\mathrm{cos}}^{-1}\left(\frac{1}{2}\right)+2{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=\frac{2\mathrm{\pi }}{3}$

#### Question 12:

Write the range of tan−1x.

#### Answer:

The range of  ${\mathrm{tan}}^{-1}x$ is$\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$.

#### Question 13:

Write the value of cos−1 (cos 1540°).

#### Answer:

We know that
${\mathrm{cos}}^{-1}\left(\mathrm{cos}x\right)=x$
Now,

#### Question 14:

Write the value of sin−1$\left(\mathrm{sin}\left(-600°\right)\right)$.

#### Answer:

We know that ${\mathrm{sin}}^{-1}\left(\mathrm{sin}x\right)=x$.
Now,

∴ ${\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(-{600}^{\circ }\right)\right\}={60}^{\circ }$

#### Question 15:

Write the value of cos .

#### Answer:

Let $y={\mathrm{sin}}^{-1}\frac{1}{3}$
Then, $\mathrm{sin}y=\frac{1}{3}$

Now, $\mathrm{cos}y=\sqrt{1-{\mathrm{sin}}^{2}y}$,

$⇒\mathrm{cos}y=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

∴ $\mathrm{cos}\left(2{\mathrm{sin}}^{-1}\frac{1}{3}\right)=\frac{7}{9}$

#### Question 16:

Write the value of sin1 (sin 1550°).

#### Answer:

We know that ${\mathrm{sin}}^{-1}\left(\mathrm{sin}x\right)=x$.
Now,

∴ ${\mathrm{sin}}^{-1}\left(\mathrm{sin}{1550}^{\circ }\right)={70}^{\circ }$

#### Question 17:

Evaluate sin $\left(\frac{1}{2}{\mathrm{cos}}^{-1}\frac{4}{5}\right)$.

#### Answer:

We know that
${\mathrm{cos}}^{-1}x=2{\mathrm{tan}}^{-1}\sqrt{\frac{1-x}{1+x}}\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{-1}x={\mathrm{sin}}^{-1}\frac{x}{\sqrt{1+{x}^{2}}}$

∴ $\mathrm{sin}\left(\frac{1}{2}{\mathrm{cos}}^{-1}\frac{4}{5}\right)=\frac{1}{\sqrt{10}}$

#### Question 18:

Evaluate sin $\left({\mathrm{tan}}^{-1}\frac{3}{4}\right)$.

#### Answer:

We know that
${\mathrm{tan}}^{-1}x={\mathrm{sin}}^{-1}\frac{x}{\sqrt{1+{x}^{2}}}$

â€‹∴ $\mathrm{sin}\left({\mathrm{tan}}^{-1}\frac{3}{4}\right)=\frac{3}{5}$

#### Question 19:

Write the value of cos−1$\left(\mathrm{tan}\frac{3\mathrm{\pi }}{4}\right)$.

#### Answer:

We have

∴ ${\mathrm{cos}}^{-1}\left(\mathrm{tan}\frac{3\mathrm{\pi }}{4}\right)=\mathrm{\pi }$

#### Question 20:

Write the value of cos .

#### Answer:

We have, cos = $\mathrm{cos}\left(2×\frac{\mathrm{\pi }}{6}\right)=\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}$

#### Question 21:

Write the value of cos1 (cos 350°) − sin−1 (sin 350°)

#### Answer:

We have

∴ ${\mathrm{cos}}^{-1}\left(\mathrm{cos}{350}^{\circ }\right)-{\mathrm{sin}}^{-1}\left(\mathrm{sin}{350}^{\circ }\right)={20}^{\circ }$

#### Question 22:

Write the value of cos2$\left(\frac{1}{2}{\mathrm{cos}}^{-1}\frac{3}{5}\right)$.

#### Answer:

Now,

${\mathrm{cos}}^{2}\left(\frac{1}{2}{\mathrm{cos}}^{-1}\frac{3}{5}\right)=\frac{4}{5}$

#### Question 23:

If tan−1x + tan−1y = $\frac{\mathrm{\pi }}{4}$, then write the value of x + y + xy.

#### Answer:

We know that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$.
Now,

∴ $x+y+xy=1$

#### Question 24:

Write the value of cos−1 (cos 6).

#### Answer:

We know that ${\mathrm{cos}}^{-1}\left(\mathrm{cos}x\right)=x$.
Now,

#### Question 25:

Write the value of sin−1$\left(\mathrm{cos}\frac{\mathrm{\pi }}{9}\right)$.

Consider,

∴

#### Question 26:

Write the value of sin $\left\{\frac{\mathrm{\pi }}{3}-{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right\}$.

#### Answer:

We have
$\mathrm{sin}\left\{\frac{\mathrm{\pi }}{3}-{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right\}=\mathrm{sin}\left\{\frac{\mathrm{\pi }}{3}-\left(-\frac{\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left\{\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}=1$

∴

#### Question 27:

Write the value of tan1$\left\{\mathrm{tan}\left(\frac{15\mathrm{\pi }}{4}\right)\right\}$.

We have

∴

#### Question 28:

Write the value of .

#### Question 29:

Write the value of ${\mathrm{tan}}^{-1}\frac{a}{b}-{\mathrm{tan}}^{-1}\left(\frac{a-b}{a+b}\right)$.

#### Answer:

We know that ${\mathrm{tan}}^{-1}x-{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x-y}{1+xy}\right)$.
Now,

∴ ${\mathrm{tan}}^{-1}\frac{a}{b}-{\mathrm{tan}}^{-1}\left(\frac{a-b}{a+b}\right)=\frac{\mathrm{\pi }}{4}$

#### Question 30:

Write the value of cos−1$\left(\mathrm{cos}\frac{5\mathrm{\pi }}{4}\right)$.

#### Answer:

${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{5\mathrm{\pi }}{4}\right)\ne \frac{5\mathrm{\pi }}{4}$ as $\frac{5\mathrm{\pi }}{4}$ does not lie between .
We have

${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{5\mathrm{\pi }}{4}\right)={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(2\mathrm{\pi }-\frac{3\mathrm{\pi }}{4}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{3\mathrm{\pi }}{4}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{3\mathrm{\pi }}{4}$

Show that .

We have

#### Question 32:

Evaluate: ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{3\mathrm{\pi }}{5}\right)$.

#### Answer:

We know that  ${\mathrm{sin}}^{-1}\left(\mathrm{sin}x\right)=x$.
We have

∴ ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{3\pi }{5}\right)=\frac{2\pi }{5}$

#### Question 33:

If ${\mathrm{tan}}^{-1}\left(\sqrt{3}\right)+{\mathrm{cot}}^{-1}x=\frac{\mathrm{\pi }}{2},$ find x.

#### Answer:

We know that ${\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}x=\frac{\mathrm{\pi }}{2}$.
We have
${\mathrm{tan}}^{-1}\left(\sqrt{3}\right)+{\mathrm{cot}}^{-1}x=\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\sqrt{3}\right)=\frac{\mathrm{\pi }}{2}-{\mathrm{cot}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\sqrt{3}\right)={\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒x=\sqrt{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$x=\sqrt{3}$

#### Question 34:

If ${\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)+{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2},$ then find x.

#### Answer:

We know that ${\mathrm{sin}}^{-1}x+{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2}$.
We have
${\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)+{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)=\frac{\mathrm{\pi }}{2}-{\mathrm{cos}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)={\mathrm{sin}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

∴ $x=\frac{1}{3}$

#### Question 35:

Write the value of ${\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)-{\mathrm{cos}}^{-1}\left(-\frac{1}{3}\right)$.

#### Answer:

We know that ${\mathrm{sin}}^{-1}x+{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2}$ and ${\mathrm{cos}}^{-1}\left(-x\right)=\mathrm{\pi }-{\mathrm{cos}}^{-1}x$.

∴ ${\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)-{\mathrm{cos}}^{-1}\left(-\frac{1}{3}\right)=-\frac{\mathrm{\pi }}{2}$

#### Question 36:

If 4 sin−1x + cos−1x = π, then what is the value of x?

#### Answer:

We know that ${\mathrm{sin}}^{-1}x+{\mathrm{cos}}^{-1}x=\frac{\pi }{2}$

∴ $x=\frac{1}{2}$

#### Question 37:

If x < 0, y < 0 such that xy = 1, then write the value of tan1x + tan−1y.

#### Answer:

We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

such that
$xy=1$

Let x = $-$a and y = $-$b where both a and b are positive.

#### Question 38:

What is the principal value of ${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$?

#### Answer:

Let $y={\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
Then,
$\mathrm{sin}y=-\frac{\sqrt{3}}{2}=\mathrm{sin}\left(-\frac{\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}y=-\frac{\mathrm{\pi }}{3}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Here, $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is the range of the principal value branch of inverse sine function.

∴ ${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\mathrm{\pi }}{3}$

#### Question 39:

Write the principal value of ${\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)$

#### Answer:

Let $y={\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)$
Then,
$\mathrm{sin}y=-\frac{1}{2}=\mathrm{sin}\left(-\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}y=-\frac{\mathrm{\pi }}{6}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Here, $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ is the range of the principal value branch of the inverse sine function.

∴ ${\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)=-\frac{\mathrm{\pi }}{6}$

#### Question 40:

Write the principal value of ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)+{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{2\mathrm{\pi }}{3}\right)$

#### Answer:

∴ ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)+{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{2\mathrm{\pi }}{3}\right)=\mathrm{\pi }$

#### Question 41:

Write the value of $\mathrm{tan}\left(2{\mathrm{tan}}^{-1}\frac{1}{5}\right)$

#### Answer:

$\mathrm{tan}\left(2{\mathrm{tan}}^{-1}\frac{1}{5}\right)=\mathrm{tan}\left[{\mathrm{tan}}^{-1}\frac{2×\frac{1}{5}}{1-{\left(\frac{1}{5}\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{\frac{2}{5}}{\frac{24}{25}}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{5}{12}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{12}$

#### Question 42:

Write.the principal value of ${\mathrm{tan}}^{-1}1+{\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)$

#### Answer:

${\mathrm{tan}}^{-1}1+{\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)={\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{\mathrm{\pi }}{4}\right)+{\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}+\frac{2\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}=\frac{11\mathrm{\pi }}{12}$

#### Question 43:

Write the value of ${\mathrm{tan}}^{-1}\left\{2\mathrm{sin}\left(2{\mathrm{cos}}^{-1}\frac{\sqrt{3}}{2}\right)\right\}$

#### Answer:

${\mathrm{tan}}^{-1}\left\{2\mathrm{sin}\left(2{\mathrm{cos}}^{-1}\frac{\sqrt{3}}{2}\right)\right\}={\mathrm{tan}}^{-1}\left\{2\mathrm{sin}\left[{\mathrm{cos}}^{-1}2{\left(\frac{\sqrt{3}}{2}\right)}^{2}-1\right]\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left[2\mathrm{sin}\left({\mathrm{cos}}^{-1}\frac{1}{2}\right)\right]$

#### Question 44:

Write the principal value of ${\mathrm{tan}}^{-1}\sqrt{3}+{\mathrm{cot}}^{-1}\sqrt{3}$

#### Answer:

We know ${\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}x=\frac{\mathrm{\pi }}{2}$
$\therefore {\mathrm{tan}}^{-1}\sqrt{3}+{\mathrm{cot}}^{-1}\sqrt{3}=\frac{\mathrm{\pi }}{2}$

#### Question 45:

Write the principal value of ${\mathrm{cos}}^{-1}\left(\mathrm{cos}680°\right)$

#### Answer:

${\mathrm{cos}}^{-1}\left(\mathrm{cos}680°\right)={\mathrm{cos}}^{-1}\left[\mathrm{cos}\left(720°-680°\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left(\mathrm{cos}40°\right)\phantom{\rule{0ex}{0ex}}=40°$

#### Question 46:

Write the value of ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{3\mathrm{\pi }}{5}\right)$

#### Answer:

${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{3\mathrm{\pi }}{5}\right)={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(\mathrm{\pi }-\frac{2\mathrm{\pi }}{5}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{2\mathrm{\pi }}{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{\pi }}{5}$

#### Question 47:

Write the value of ${\mathrm{sec}}^{-1}\left(\frac{1}{2}\right)$.

#### Answer:

The value of ${\mathrm{sec}}^{-1}\left(\frac{1}{2}\right)$ is undefined as it is outside the range i.e., R – (–1, 1) .

#### Question 48:

Write the value of ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{14\mathrm{\pi }}{3}\right)$

#### Answer:

${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{14\mathrm{\pi }}{3}\right)={\mathrm{cos}}^{-1}\left[\mathrm{cos}\left(4\mathrm{\pi }+\frac{2\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{\pi }}{3}$

#### Question 49:

Write the value of

We have

Now,

#### Question 50:

Wnte the value of the expression

#### Question 51:

Write the principal value of ${\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{1}{2}\right)\right\}$

#### Answer:

${\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{1}{2}\right)\right\}={\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left[{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{3}\right)\right]\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left[\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left[\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}$

#### Question 52:

The set of values of ${\mathrm{cosec}}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

#### Answer:

The value of ${\mathrm{cosec}}^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is undefined as it is outside the range i.e., R – (–1, 1) .

#### Question 53:

Write the value of  ${\mathrm{tan}}^{-1}\left(\frac{1}{x}\right)$ for x < 0 in terms of ${\mathrm{cot}}^{-1}x$

#### Question 54:

Write the value of  ${\mathrm{cot}}^{-1}\left(-x\right)$ for all $x\in R$  in terms of ${\mathrm{cot}}^{-1}x$

#### Answer:

We know that ${\mathrm{cot}}^{-1}\left(-x\right)=\mathrm{\pi }-{\mathrm{cot}}^{-1}\left(x\right)$
Therefore, the value of  ${\mathrm{cot}}^{-1}\left(-x\right)$ for all $x\in R$  in terms of ${\mathrm{cot}}^{-1}x$ is $\mathrm{\pi }-{\mathrm{cot}}^{-1}\left(x\right)$.

#### Question 55:

Wnte the value of

#### Question 56:

If $\mathrm{cos}\left({\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}\sqrt{3}\right)=0$, find the value of x.

#### Question 57:

Find the value of $2{\mathrm{sec}}^{-1}2+{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)$

#### Answer:

$2{\mathrm{sec}}^{-1}2+{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=2{\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{\mathrm{\pi }}{3}\right)+{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=2×\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}=\frac{5\mathrm{\pi }}{6}$

#### Question 58:

If $\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{2}{5}+{\mathrm{cos}}^{-1}x\right)=0$, find the value of x.

#### Question 59:

Find the value of ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{13\mathrm{\pi }}{6}\right)$

#### Answer:

${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{13\mathrm{\pi }}{6}\right)={\mathrm{cos}}^{-1}\left[\mathrm{cos}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left[\mathrm{cos}\left(\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}$

#### Question 60:

Find the value of ${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{9\mathrm{\pi }}{8}\right)$

#### Answer:

${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{9\mathrm{\pi }}{8}\right)={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{8}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{8}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{8}$

#### Question 1:

If ${\mathrm{tan}}^{-1}\left(\frac{\sqrt{1+{x}^{2}}-\sqrt{1-{x}^{2}}}{\sqrt{1+{x}^{2}}+\sqrt{1-{x}^{2}}}\right)$ = α, then x2 =
(a) sin 2 α
(b) sin α
(c) cos 2 α
(d) cos α

(a) sin 2α

#### Question 2:

The value of tan $\left\{{\mathrm{cos}}^{-1}\frac{1}{5\sqrt{2}}-{\mathrm{sin}}^{-1}\frac{4}{\sqrt{17}}\right\}$ is
(a) $\frac{\sqrt{29}}{3}$

(b) $\frac{29}{3}$

(c) $\frac{\sqrt{3}}{29}$

(d) $\frac{3}{29}$

#### Answer:

(d) $\frac{3}{29}$

#### Question 3:

2 tan−1 {cosec (tan−1x) − tan (cot1x)} is equal to
(a) cot−1x

(b) cot−1$\frac{1}{x}$

(c) tan−1x

(d) none of these

#### Answer:

(c) tan−1x

Let ${\mathrm{tan}}^{-1}x=y$
So, $x=\mathrm{tan}y$

If
(a) sin2 α
(b) cos2 α
(c) tan2 α
(d) cot2 α

#### Answer:

(a) sin2 α

We know that ${\mathrm{cos}}^{-1}x+{\mathrm{cos}}^{-1}y={\mathrm{cos}}^{-1}\left(xy-\sqrt{1-{x}^{2}}\sqrt{1-{y}^{2}}\right)$.

#### Question 5:

The positive integral solution of the equation
${\mathrm{tan}}^{-1}x+{\mathrm{cos}}^{-1}\frac{y}{\sqrt{1+{y}^{2}}}={\mathrm{sin}}^{-1}\frac{3}{\sqrt{10}}\mathrm{is}$
(a) x = 1, y = 2
(b) x = 2, y = 1
(c) x = 3, y = 2
(d) x = −2, y = −1.

(a) x = 1, y = 2

#### Question 6:

If sin−1x − cos−1x = $\frac{\mathrm{\pi }}{6}$, then x =
(a) $\frac{1}{2}$

(b) $\frac{\sqrt{3}}{2}$

(c) $-\frac{1}{2}$

(d) none of these

#### Answer:

(b) $\frac{\sqrt{3}}{2}$
We know that ${\mathrm{sin}}^{-1}x+{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2}$.

#### Question 7:

sin $\left[{\mathrm{cot}}^{-1}\left\{\mathrm{tan}\left({\mathrm{cos}}^{-1}x\right)\right\}\right]$ is equal to
(a) x

(b) $\sqrt{1-{x}^{2}}$

(c) $\frac{1}{x}$

(d) none of these

#### Answer:

(a) x

Let ${\mathrm{cos}}^{-1}x=y$

Then,

#### Question 8:

The number of solutions of the equation
${\mathrm{tan}}^{-1}2x+{\mathrm{tan}}^{-1}3x=\frac{\mathrm{\pi }}{4}$ is
(a) 2
(b) 3
(c) 1
(d) none of these

#### Answer:

(a) 2
We know that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$.

Therefore, there are two solutions.

#### Question 9:

If α = , then
(a) 4 α = 3 β
(b) 3 α = 4 β
(c) α − β = $\frac{7\mathrm{\pi }}{12}$
(d) none of these

#### Answer:

(a) 4 α = 3 β

We know that ${\mathrm{tan}}^{-1}\left(\mathrm{tan}x\right)=x$.

and
$\beta ={\mathrm{tan}}^{-1}\left\{-\mathrm{tan}\left(\frac{2\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{-\mathrm{tan}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{\mathrm{tan}\left(\frac{\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}$

∴

#### Question 10:

The number of real solutions of the equation
is
(a) 0
(b) 1
(c) 2
(d) infinite

(c) 2

#### Question 11:

If x < 0, y < 0 such that xy = 1, then tan−1x + tan−1y equals
(a) $\frac{\mathrm{\pi }}{2}$

(b) $-\frac{\mathrm{\pi }}{2}$

(c) − π

(d) none of these

#### Answer:

(b) $-\frac{\mathrm{\pi }}{2}$
We know that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$.
such that
$xy=1$

Let x = $-$a and y = $-$b, where a and b both are positive.

(a)

(b)

(c) tan θ

(d) cot θ

#### Answer:

(a)

Let $y=\sqrt{\mathrm{tan}\theta }$
Then,

#### Question 13:

(a) 36
(b) 36 − 36 cos θ
(c) 18 − 18 cos θ
(d) 18 + 18 cos θ

#### Answer:

(c) 18 − 18 cosθ

We know
${\mathrm{cos}}^{-1}x+{\mathrm{cos}}^{-1}y={\mathrm{cos}}^{-1}\left(xy-\sqrt{1-{x}^{2}}\sqrt{1-{y}^{2}}\right)$

Squaring both the sides, we get

#### Question 14:

If α = then α − β =
(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{\mathrm{\pi }}{3}$

(c) $\frac{\mathrm{\pi }}{2}$

(d) $-\frac{\mathrm{\pi }}{3}$

#### Answer:

(a) $\frac{\mathrm{\pi }}{6}$

We have
α =

#### Question 15:

Let f (x) = . Then, f (8π/9) =
(a) e5π/18
(b) e13π/18
(c) e−2π/18
(d) none of these

#### Answer:

(b) e13π/18

Given: $f\left(x\right)={e}^{{\mathrm{cos}}^{-1}\left\{\mathrm{sin}\left(x+\pi }{3}\right)\right\}}$

Then,

#### Question 16:

${\mathrm{tan}}^{-1}\frac{1}{11}+{\mathrm{tan}}^{-1}\frac{2}{11}$ is equal to
(a) 0
(b) 1/2
(c) − 1
(d) none of these

#### Answer:

(d) none of these

We know that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$.
Now,
${\mathrm{tan}}^{-1}\frac{1}{11}+{\mathrm{tan}}^{-1}\frac{2}{11}={\mathrm{tan}}^{-1}\left(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{1}{11}\frac{2}{11}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{\frac{3}{11}}{\frac{121-2}{121}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{\frac{3}{11}}{\frac{119}{121}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{33}{119}\right)\phantom{\rule{0ex}{0ex}}=0.27$

#### Question 17:

If ${\mathrm{cos}}^{-1}\frac{x}{2}+{\mathrm{cos}}^{-1}\frac{y}{3}=\mathrm{\theta },$ then 9x2 − 12xy cos θ + 4y2 is equal to
(a) 36
(b) −36 sin2 θ
(c) 36 sin2 θ
(d) 36 cos2 θ

#### Answer:

(c) 36 sin2 θ

We know
${\mathrm{cos}}^{-1}x+{\mathrm{cos}}^{-1}y={\mathrm{cos}}^{-1}\left[xy-\sqrt{1-{x}^{2}}\sqrt{1-{y}^{2}}\right]$

Now,

#### Question 18:

If tan−1 3 + tan−1x = tan−1 8, then x =
(a) 5
(b) 1/5
(c) 5/14
(d) 14/5

#### Answer:

(b) $\frac{1}{5}$
We know that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\frac{x+y}{1-xy}$.
Now,
${\mathrm{tan}}^{-1}3+{\mathrm{tan}}^{-1}x={\mathrm{tan}}^{-1}8\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{3+x}{1-3x}\right)={\mathrm{tan}}^{-1}8\phantom{\rule{0ex}{0ex}}⇒\frac{3+x}{1-3x}=8\phantom{\rule{0ex}{0ex}}⇒3+x=8-24x\phantom{\rule{0ex}{0ex}}⇒3-8=-24x-x\phantom{\rule{0ex}{0ex}}⇒-5=-25x\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{25}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}$

#### Question 19:

The value of ${\mathrm{sin}}^{-1}\left(\mathrm{cos}\frac{33\mathrm{\pi }}{5}\right)$ is
(a) $\frac{3\mathrm{\pi }}{5}$

(b) $-\frac{\mathrm{\pi }}{10}$

(c) $\frac{\mathrm{\pi }}{10}$

(d) $\frac{7\mathrm{\pi }}{5}$

#### Answer:

(b) $-\frac{\mathrm{\pi }}{10}$

${\mathrm{sin}}^{-1}\left(\mathrm{cos}\frac{33\mathrm{\pi }}{5}\right)={\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left(6\mathrm{\pi }+\frac{3\mathrm{\pi }}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left(\frac{3\mathrm{\pi }}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-\frac{3\mathrm{\pi }}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}-\frac{3\mathrm{\pi }}{5}\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{10}\phantom{\rule{0ex}{0ex}}$

#### Question 20:

The value of ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{5\mathrm{\pi }}{3}\right)+{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{5\mathrm{\pi }}{3}\right)$ is
(a) $\frac{\mathrm{\pi }}{2}$

(b) $\frac{5\mathrm{\pi }}{3}$

(c) $\frac{10\mathrm{\pi }}{3}$

(d) 0

#### Answer:

(d) 0

We have
${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{5\mathrm{\pi }}{3}\right)+{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{5\mathrm{\pi }}{3}\right)={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right\}+{\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)\right\}+{\mathrm{sin}}^{-1}\left\{-\mathrm{sin}\left(\frac{\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)\right\}-{\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\frac{\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}=0$

#### Question 21:

sin is equal to
(a) $\frac{6}{25}$

(b) $\frac{24}{25}$

(c) $\frac{4}{5}$

(d) $-\frac{24}{25}$

#### Answer:

(d) $-\frac{24}{25}$

Let
Then,

Now,

#### Question 22:

If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
(a) $\frac{\mathrm{\pi }}{3}$

(b) $\frac{\mathrm{\pi }}{2}$

(c) $\frac{2\mathrm{\pi }}{3}$

(d) $-\frac{2\mathrm{\pi }}{3}$

#### Answer:

(a) $\frac{\mathrm{\pi }}{3}$

We know
${\mathrm{sin}}^{-1}\left(\mathrm{sin}x\right)=x$

Now,

#### Question 23:

If 3 is equal to
(a) $\frac{1}{\sqrt{3}}$

(b) $-\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $-\frac{\sqrt{3}}{4}$

#### Answer:

(a) $\frac{1}{\sqrt{3}}$

Let $x=\mathrm{tan}y$
Then,

#### Question 24:

If 4 cos−1x + sin−1x = π, then the value of x is
(a) $\frac{3}{2}$

(b) $\frac{1}{\sqrt{2}}$

(c) $\frac{\sqrt{3}}{2}$

(d) $\frac{2}{\sqrt{3}}$

#### Answer:

(c) $\frac{\sqrt{3}}{2}$
We know that ${\mathrm{sin}}^{-1}x+{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2}$.

$4{\mathrm{cos}}^{-1}x+{\mathrm{sin}}^{-1}x=\mathrm{\pi }\phantom{\rule{0ex}{0ex}}⇒4{\mathrm{cos}}^{-1}x+\frac{\mathrm{\pi }}{2}-{\mathrm{cos}}^{-1}x=\mathrm{\pi }\phantom{\rule{0ex}{0ex}}⇒3{\mathrm{cos}}^{-1}x=\mathrm{\pi }-\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}⇒3{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒x=\mathrm{cos}\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{\sqrt{3}}{2}$

#### Question 25:

It ${\mathrm{tan}}^{-1}\frac{x+1}{x-1}+{\mathrm{tan}}^{-1}\frac{x-1}{x}={\mathrm{tan}}^{-1}$ (−7), then the value of x is
(a) 0
(b) −2
(c) 1
(d) 2

#### Answer:

(d) 2

We know that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$.

So, we get

$\frac{2{x}^{2}-x+1}{-x+1}=-7\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-x+1=7x-7\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-8x+8=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-4x+4=0\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒x=2$

#### Question 26:

If ${\mathrm{cos}}^{-1}x>{\mathrm{sin}}^{-1}x$, then

(a) $\frac{1}{\sqrt{2}}
(b) $0\le x<\frac{1}{\sqrt{2}}$
(c)$-1\le x<\frac{1}{\sqrt{2}}$
(d) x > 0

#### Answer:

${\mathrm{cos}}^{-1}x>{\mathrm{sin}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{-1}x>\frac{\mathrm{\pi }}{2}-{\mathrm{cos}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{cos}}^{-1}x>\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{-1}x>\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒x>\mathrm{cos}\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒x>\frac{1}{\sqrt{2}}$
We know that the maximum value of cosine fuction is 1.
$\therefore \frac{1}{\sqrt{2}}
Hence, the correct answer is option(a).

#### Question 27:

In a âˆ† ABC, if C is a right angle, then
${\mathrm{tan}}^{-1}\left(\frac{a}{b+c}\right)+{\mathrm{tan}}^{-1}\left(\frac{b}{c+a}\right)=$

(a) $\frac{\mathrm{\pi }}{3}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{5\mathrm{\pi }}{2}$

(d) $\frac{\mathrm{\pi }}{6}$

#### Answer:

(b) $\frac{\mathrm{\pi }}{4}$

We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

#### Question 28:

The value of sin$\left(\frac{1}{4}{\mathrm{sin}}^{-1}\frac{\sqrt{63}}{8}\right)$ is
(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\frac{1}{2\sqrt{2}}$

(d) $\frac{1}{3\sqrt{3}}$

#### Answer:

(c) $\frac{1}{2\sqrt{2}}$

Let ${\mathrm{sin}}^{-1}\frac{\sqrt{63}}{8}=y$
Then,
$\mathrm{sin}y=\frac{\sqrt{63}}{8}\phantom{\rule{0ex}{0ex}}\mathrm{cos}y=\sqrt{1-{\mathrm{sin}}^{2}y}=\sqrt{1-\frac{63}{64}}=\frac{1}{8}$
Now, we have

#### Question 29:

(a) 7
(b) 6
(c) 5
(d) none of these

#### Answer:

(a) 7

Let $2{\mathrm{cot}}^{-1}3=y$
Then, $\mathrm{cot}\frac{y}{2}=3$

#### Question 30:

If tan−1 (cot θ) = 2 θ, then θ =

(a) $±\frac{\mathrm{\pi }}{3}$

(b) $±\frac{\mathrm{\pi }}{4}$

(c) $±\frac{\mathrm{\pi }}{6}$

(d) none of these

#### Answer:

(c) $±\frac{\mathrm{\pi }}{6}$

#### Question 31:

If , then, the value of x is

(a) 0
(b) $\frac{a}{2}$
(c) a
(d) $\frac{2a}{1-{a}^{2}}$

#### Answer:

${\mathrm{sin}}^{-1}\left(\frac{2a}{1-{a}^{2}}\right)+{\mathrm{cos}}^{-1}\left(\frac{1-{a}^{2}}{1+{a}^{2}}\right)={\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{tan}}^{-1}a+2{\mathrm{tan}}^{-1}a=2{\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒4{\mathrm{tan}}^{-1}a=2{\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{tan}}^{-1}a={\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{2a}{1-{a}^{2}}\right)={\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒x=\frac{2a}{1-{a}^{2}}$
Hence, the correct answer is option(d).

#### Question 32:

The value of  $\mathrm{sin}\left(2\left({\mathrm{tan}}^{-1}0.75\right)\right)$is equal to
(a) 0.75
(b) 1.5
(c) 0.96
(d) ${\mathrm{sin}}^{-1}1.5$

#### Answer:

$\mathrm{sin}\left(2\left({\mathrm{tan}}^{-1}0.75\right)\right)=\mathrm{sin}\left(2{\mathrm{tan}}^{-1}0.75\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{2×0.75}{1+{\left(0.75\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left({\mathrm{sin}}^{-1}0.96\right)\phantom{\rule{0ex}{0ex}}=0.96$
Hence, the correct answer is option (c).

#### Question 33:

If x > 1, then $2{\mathrm{tan}}^{-1}x+{\mathrm{sin}}^{-1}\left(\frac{2x}{1+{x}^{2}}\right)$is equal to
(a) $4{\mathrm{tan}}^{-1}x$
(b) 0
(c) $\frac{\mathrm{\pi }}{2}$
(d) $\mathrm{\pi }$

#### Answer:

Hence, the correct answer is option (a)

#### Question 34:

The domain of ${\mathrm{cos}}^{-1}\left({x}^{2}-4\right)$ is
(a) [3, 5]
(b) [−1, 1]
(c)
(d)

#### Answer:

The domain of ${\mathrm{cos}}^{-1}\left(x\right)$ is [−1, 1]

Hence, the correct answer is option (c).

#### Question 35:

The value of $\mathrm{tan}\left({\mathrm{cos}}^{-1}\frac{3}{5}+{\mathrm{tan}}^{-1}\frac{1}{4}\right)$
(a) $\frac{19}{8}$
(b) $\frac{8}{19}$
(c) $\frac{19}{12}$
(d) $\frac{3}{4}$

#### Answer:

$\mathrm{tan}\left({\mathrm{cos}}^{-1}\frac{3}{5}+{\mathrm{tan}}^{-1}\frac{1}{4}\right)=\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}+{\mathrm{tan}}^{-1}\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{\frac{4}{5}}{\frac{3}{5}}+{\mathrm{tan}}^{-1}\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{4}{3}+{\mathrm{tan}}^{-1}\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{1}{3}}\right)\phantom{\rule{0ex}{0ex}}=\frac{\frac{16+3}{12}}{\frac{2}{3}}\phantom{\rule{0ex}{0ex}}=\frac{19}{8}$
Hence, the correct answer is option (a).

#### Question 1:

Find the principal values of each of the following:

(i) ${\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
(ii) ${\mathrm{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

(iii) ${\mathrm{tan}}^{-1}\left(\mathrm{cos}\frac{\mathrm{\pi }}{2}\right)$

(iv) ${\mathrm{tan}}^{-1}\left(2\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)$

#### Answer:

(i) Let ${\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$
Then,
$\mathrm{tan}y=\frac{1}{\sqrt{3}}$
We know that the range of the principal value branch is $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$.
Thus,
$\mathrm{tan}y=\frac{1}{\sqrt{3}}=\mathrm{tan}\left(\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{\mathrm{\pi }}{6}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$
Hence, the principal value of .

(ii) We have
Let ${\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$
Then,
$\mathrm{tan}y=\frac{1}{\sqrt{3}}$
We know that the range of the principal value branch is $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$.
Thus,
$\mathrm{tan}y=\frac{1}{\sqrt{3}}=\mathrm{tan}\left(\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-{\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)\phantom{\rule{0ex}{0ex}}=-y\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{6}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$
Hence, the principal value of .

(iii) Let ${\mathrm{tan}}^{-1}\left(\mathrm{cos}\frac{\mathrm{\pi }}{2}\right)=y$
Then,
$\mathrm{tan}y=\mathrm{cos}\frac{\mathrm{\pi }}{2}$
We know that the range of the principal value branch is $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$.
Thus,
$\mathrm{tan}y=\mathrm{cos}\frac{\mathrm{\pi }}{2}=0=\mathrm{tan}\left(0\right)\phantom{\rule{0ex}{0ex}}⇒y=0\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$
Hence, the principal value of .

(iv)  Let ${\mathrm{tan}}^{-1}\left(2\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)=y$
Then,
$\mathrm{tan}y=2\mathrm{cos}\frac{2\mathrm{\pi }}{3}$
We know that the range of the principal value branch is $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$.
Thus,
$\mathrm{tan}y=2\mathrm{cos}\frac{2\mathrm{\pi }}{3}=2×\frac{-1}{2}=-1=\mathrm{tan}\left(-\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}⇒y=-\frac{\mathrm{\pi }}{4}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$
Hence, the principal value of .

#### Question 2:

For the principal values, evaluate each of the following:

(ii) ${\mathrm{tan}}^{-1}\left\{2\mathrm{sin}\left(4{\mathrm{cos}}^{-1}\frac{\sqrt{3}}{2}\right)\right\}$

#### Answer:

$\therefore {\mathrm{tan}}^{-1}\left(-1\right)+{\mathrm{cos}}^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{\mathrm{\pi }}{2}$

(ii)
${\mathrm{tan}}^{-1}\left\{2\mathrm{sin}\left(4{\mathrm{cos}}^{-1}\frac{\sqrt{3}}{2}\right)\right\}={\mathrm{tan}}^{-1}\left\{2\mathrm{sin}\left[4{\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{\mathrm{\pi }}{6}\right)\right]\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{2\mathrm{sin}\left[4×\frac{\mathrm{\pi }}{6}\right]\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(2\mathrm{sin}\frac{2\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left[2×\left(\frac{\sqrt{3}}{2}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}$

#### Question 3:

Evaluate each of the following:

(ii) ${\mathrm{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\mathrm{tan}}^{-1}\left(-\sqrt{3}\right)+{\mathrm{tan}}^{-1}\left(\mathrm{sin}\left(-\frac{\mathrm{\pi }}{2}\right)\right)$

(iii) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{5\mathrm{\pi }}{6}\right)+{\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{13\mathrm{\pi }}{6}\right)\right\}$

#### Answer:

(i)  Let ${\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)=y$
Then,
$\mathrm{sin}y=-\frac{1}{2}$
We know that the range of the principal value branch is $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$.
Thus,
$\mathrm{sin}y=-\frac{1}{2}=\mathrm{sin}\left(-\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}⇒y=-\frac{\mathrm{\pi }}{6}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$
Now,
Let ${\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)=z$
Then,
$\mathrm{cos}z=-\frac{1}{2}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]$.
Thus,
$\mathrm{cos}z=-\frac{1}{2}=\mathrm{cos}\left(\frac{2\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}⇒z=\frac{2\mathrm{\pi }}{3}\in \left[0,\mathrm{\pi }\right]$
So,

$\therefore {\mathrm{tan}}^{-1}1+{\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)+{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=\frac{3\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}$

(ii)
${\mathrm{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\mathrm{tan}}^{-1}\left(-\sqrt{3}\right)+{\mathrm{tan}}^{-1}\left(\mathrm{sin}\left(-\frac{\mathrm{\pi }}{2}\right)\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\mathrm{tan}}^{-1}\left(-\sqrt{3}\right)+{\mathrm{tan}}^{-1}\left(-\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}\right)\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\mathrm{tan}}^{-1}\left(-\sqrt{3}\right)+{\mathrm{tan}}^{-1}\left(-1\right)\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)-{\mathrm{tan}}^{-1}\left(\sqrt{3}\right)-{\mathrm{tan}}^{-1}\left(1\right)\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{\mathrm{\pi }}{6}\right)-{\mathrm{tan}}^{-1}\left(\frac{\mathrm{\pi }}{3}\right)-{\mathrm{tan}}^{-1}\left(\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}=-\frac{3\mathrm{\pi }}{4}$

(iii)
${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{5\mathrm{\pi }}{6}\right)+{\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{13\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{\mathrm{tan}\left(\mathrm{\pi }-\frac{5\mathrm{\pi }}{6}\right)\right\}+{\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{-\mathrm{tan}\left(\frac{\mathrm{\pi }}{6}\right)\right\}+{\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}\left\{\mathrm{tan}\left(\frac{\mathrm{\pi }}{6}\right)\right\}+{\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}=0$

#### Question 1:

Find the principal values of each of the following:

(i) ${\mathrm{sec}}^{-1}\left(-\sqrt{2}\right)$
(ii) ${\mathrm{sec}}^{-1}\left(2\right)$
(iii) ${\mathrm{sec}}^{-1}\left(2\mathrm{sin}\frac{3\mathrm{\pi }}{4}\right)$
(iv) ${\mathrm{sec}}^{-1}\left(2\mathrm{tan}\frac{3\mathrm{\pi }}{4}\right)$

#### Answer:

(i)  Let ${\mathrm{sec}}^{-1}\left(-\sqrt{2}\right)=y$
Then,
$\mathrm{sec}y=-\sqrt{2}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}$.
Thus,

Hence, the principal value of .

(ii) Let
${\mathrm{sec}}^{-1}\left(2\right)=y$
Then,
$\mathrm{sec}y=2$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}$.
Thus,

Hence, the principal value of .

(iii)
Let
${\mathrm{sec}}^{-1}\left(2\mathrm{sin}\frac{3\mathrm{\pi }}{4}\right)=y$
Then,
$\mathrm{sec}y=2\mathrm{sin}\frac{3\mathrm{\pi }}{4}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}$.
Thus,
$\mathrm{sec}y=2\mathrm{sin}\frac{3\mathrm{\pi }}{4}=2×\frac{1}{\sqrt{2}}=\sqrt{2}=\mathrm{sec}\left(\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{\mathrm{\pi }}{4}\in \left[0,\mathrm{\pi }\right]$
Hence, the principal value of .
(iv)
Let ${\mathrm{sec}}^{-1}\left(2\mathrm{tan}\frac{3\mathrm{\pi }}{4}\right)=y$
Then,
$\mathrm{sec}y=2\mathrm{tan}\frac{3\mathrm{\pi }}{4}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}$.
Thus,
$\mathrm{sec}y=2\mathrm{tan}\frac{3\mathrm{\pi }}{4}=2×\left(-1\right)=-2=\mathrm{sec}\left(\frac{2\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{2\mathrm{\pi }}{3}\in \left[0,\mathrm{\pi }\right]$
Hence, the principal value of .

#### Question 2:

For the principal values, evaluate the following:

(i) ${\mathrm{tan}}^{-1}\sqrt{3}-{\mathrm{sec}}^{-1}\left(-2\right)$

(ii) ${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)-2{\mathrm{sec}}^{-1}\left(2\mathrm{tan}\frac{\mathrm{\pi }}{6}\right)$

#### Answer:

(i)
${\mathrm{tan}}^{-1}\sqrt{3}-{\mathrm{sec}}^{-1}\left(-2\right)={\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{\mathrm{\pi }}{3}\right)-{\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{2\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}-\frac{2\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{3}$

(ii)
${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)-2{\mathrm{sec}}^{-1}\left(2\mathrm{tan}\frac{\mathrm{\pi }}{6}\right)=-{\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)-2{\mathrm{sec}}^{-1}\left(2×\frac{1}{\sqrt{3}}\right)\phantom{\rule{0ex}{0ex}}=-{\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)-2{\mathrm{sec}}^{-1}\left(\frac{2}{\sqrt{3}}\right)\phantom{\rule{0ex}{0ex}}=-{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{3}\right)-2{\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}=-\frac{2\mathrm{\pi }}{3}$

#### Question 3:

Find the domain of
(i) ${\mathrm{sec}}^{-1}\left(3x-1\right)$
(ii) ${\mathrm{sec}}^{-1}x-{\mathrm{tan}}^{-1}x$

#### Answer:

(ii)

Let f(x) = g(x) − h(x), where g(x)=cotx and h(x)=cot1x
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is [0,  π/2) â‹ƒ [ π, 3π/2)
The domain of h(x) is
Therfore, the intersection of g(x) and h(x) is R − { nπ, n â‹µ Z}

#### Question 1:

â€‹Find the principal values of each of the following:

(i)  ${\mathrm{cosec}}^{-1}\left(-\sqrt{2}\right)$
(ii) ${\mathrm{cosec}}^{-1}\left(-2\right)$
(iii) ${\mathrm{cosec}}^{-1}\left(\frac{2}{\sqrt{3}}\right)$
(iv) ${\mathrm{cosec}}^{-1}\left(2\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)$

#### Answer:

(i)  Let ${\mathrm{cosec}}^{-1}\left(-\sqrt{2}\right)=y$
Then,
$\mathrm{cosec}y=-\sqrt{2}$
We know that the range of the principal value branch is .
Thus,

Hence, the principal value of .

(ii)
Let
${\mathrm{cosec}}^{-1}\left(-2\right)=y$
Then,
$\mathrm{cosec}y=-2$
We know that the range of the principal value branch is .
Thus,

Hence, the principal value of .

(iii) Let ${\mathrm{cosec}}^{-1}\left(\frac{2}{\sqrt{3}}\right)=y$
Then,
$\mathrm{cosec}y=\frac{2}{\sqrt{3}}$
We know that the range of the principal value branch is .
Thus,

Hence, the principal value of .

(iv)
Let
${\mathrm{cosec}}^{-1}\left(2\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)=y$
Then,
$\mathrm{cosec}y=2\mathrm{cos}\frac{2\mathrm{\pi }}{3}$
We know that the range of the principal value branch is .
Thus,

Hence, the principal value of .

#### Question 2:

Find the set of values of ${\mathrm{cosec}}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

#### Answer:

The value of ${\mathrm{cosec}}^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is undefined as it is outside the range i.e., R – (–1, 1) .

#### Question 3:

For the principal values, evaluate the following:

(i) ${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)+{\mathrm{cosec}}^{-1}\left(-\frac{2}{\sqrt{3}}\right)$
(ii) ${\mathrm{sec}}^{-1}\left(\sqrt{2}\right)+2{\mathrm{cosec}}^{-1}\left(-\sqrt{2}\right)$
(iii) ${\mathrm{sin}}^{-1}\left[\mathrm{cos}\left\{2{\mathrm{cosec}}^{-1}\left(-2\right)\right\}\right]$
(iv) ${\mathrm{cosec}}^{-1}\left(2\mathrm{tan}\frac{11\mathrm{\pi }}{6}\right)$

#### Answer:

(i)
${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)+{\mathrm{cosec}}^{-1}\left(-\frac{2}{\sqrt{3}}\right)=-{\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)+{\mathrm{cosec}}^{-1}\left(-\frac{2}{\sqrt{3}}\right)\phantom{\rule{0ex}{0ex}}=-{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{3}\right)+{\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(-\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}=-\frac{2\mathrm{\pi }}{3}$
(ii)
${\mathrm{sec}}^{-1}\left(\sqrt{2}\right)+2{\mathrm{cosec}}^{-1}\left(-\sqrt{2}\right)={\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{\mathrm{\pi }}{4}\right)+2{\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(-\frac{\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}-2×\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{4}$
(iii)

${\mathrm{sin}}^{-1}\left[\mathrm{cos}\left\{2{\mathrm{cosec}}^{-1}\left(-2\right)\right\}\right]={\mathrm{sin}}^{-1}\left[\mathrm{cos}\left\{2{\mathrm{cosec}}^{-1}\left(\mathrm{cosec}-\frac{\mathrm{\pi }}{6}\right)\right\}\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left[\mathrm{cos}\left\{-\frac{\mathrm{\pi }}{3}\right\}\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left[\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}$
(iv)
${\mathrm{cosec}}^{-1}\left(2\mathrm{tan}\frac{11\mathrm{\pi }}{6}\right)={\mathrm{cosec}}^{-1}\left[2×\left(-\frac{1}{\sqrt{3}}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cosec}}^{-1}\left[-\frac{2}{\sqrt{3}}\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(-\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{3}$

#### Question 1:

Find the principal values of each of the following:

(i) ${\mathrm{cot}}^{-1}\left(-\sqrt{3}\right)$
(ii) ${\mathrm{cot}}^{-1}\left(\sqrt{3}\right)$
(iii) ${\mathrm{cot}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)$
(iv) ${\mathrm{cot}}^{-1}\left(\mathrm{tan}\frac{3\mathrm{\pi }}{4}\right)$

#### Answer:

(i)  Let ${\mathrm{cot}}^{-1}\left(-\sqrt{3}\right)=y$
Then,
$\mathrm{cot}y=-\sqrt{3}$
We know that the range of the principal value branch is $\left(0,\mathrm{\pi }\right)$.
Thus,
$\mathrm{cot}y=-\sqrt{3}=\mathrm{cot}\left(\frac{5\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{5\mathrm{\pi }}{6}\in \left(0,\mathrm{\pi }\right)$
Hence, the principal value of .

(ii) Let
${\mathrm{cot}}^{-1}\left(\sqrt{3}\right)=y$
Then,
$\mathrm{cot}y=\sqrt{3}$
We know that the range of the principal value branch is $\left(0,\mathrm{\pi }\right)$.
Thus,
$\mathrm{cot}y=\sqrt{3}=\mathrm{cot}\left(\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{\mathrm{\pi }}{6}\in \left(0,\mathrm{\pi }\right)$
Hence, the principal value of .
(iii) Let ${\mathrm{cot}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)=y$
Then,
$\mathrm{cot}y=-\frac{1}{\sqrt{3}}$
We know that the range of the principal value branch is $\left(0,\mathrm{\pi }\right)$.
Thus,
$\mathrm{cot}y=-\frac{1}{\sqrt{3}}=\mathrm{cot}\left(\frac{2\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{2\mathrm{\pi }}{3}\in \left(0,\mathrm{\pi }\right)$
Hence, the principal value of .
(iv)
Let ${\mathrm{cot}}^{-1}\left(\mathrm{tan}\frac{3\mathrm{\pi }}{4}\right)=y$
Then,
$\mathrm{cot}y=\mathrm{tan}\frac{3\mathrm{\pi }}{4}$
We know that the range of the principal value branch is $\left(0,\mathrm{\pi }\right)$.
Thus,
$\mathrm{cot}y=\mathrm{tan}\frac{3\mathrm{\pi }}{4}=-1=\mathrm{cot}\left(\frac{3\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{3\mathrm{\pi }}{4}\in \left(0,\mathrm{\pi }\right)$
Hence, the principal value of .

#### Question 2:

Find the domain of $f\left(x\right)=\mathrm{cot}x+{\mathrm{cot}}^{-1}x$

#### Answer:

Let f(x) = g(x) + h(x), where
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is R − { nπ, n â‹µ Z}
The domain of h(x) is (0, π )
Therfore, the intersection of g(x) and h(x) is R − { nπ, n â‹µ Z}

#### Question 3:

Evaluate each of the following:

(i) ${\mathrm{cot}}^{-1}\frac{1}{\sqrt{3}}-{\mathrm{cosec}}^{-1}\left(-2\right)+{\mathrm{sec}}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

(ii) ${\mathrm{cot}}^{-1}\left\{2\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{\sqrt{3}}{2}\right)\right\}$

(iii) ${\mathrm{cosec}}^{-1}\left(-\frac{2}{\sqrt{3}}\right)+2{\mathrm{cot}}^{-1}\left(-1\right)$

(iv) ${\mathrm{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\mathrm{cot}}^{-1}\left(\frac{1}{\sqrt{3}}\right)+{\mathrm{tan}}^{-1}\left(\mathrm{sin}\left(-\frac{\mathrm{\pi }}{2}\right)\right)$

#### Answer:

(i)
${\mathrm{cot}}^{-1}\frac{1}{\sqrt{3}}-{\mathrm{cosec}}^{-1}\left(-2\right)+{\mathrm{sec}}^{-1}\left(\frac{2}{\sqrt{3}}\right)={\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{3}\right)-{\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(-\frac{\mathrm{\pi }}{6}\right)\right]+{\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{\pi }}{3}$

(ii)
${\mathrm{cot}}^{-1}\left\{2\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{\sqrt{3}}{2}\right)\right\}={\mathrm{cot}}^{-1}\left\{2\mathrm{cos}\left[{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{3}\right)\right]\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(2\mathrm{cos}\frac{\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(2×\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(1\right)\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(\mathrm{tan}\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$

(iii)
${\mathrm{cosec}}^{-1}\left(-\frac{2}{\sqrt{3}}\right)+2{\mathrm{cot}}^{-1}\left(-1\right)={\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(-\frac{\mathrm{\pi }}{3}\right)\right]+2{\mathrm{cot}}^{-1}\left[\mathrm{cot}\left(\frac{3\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{3}+2×\frac{3\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{3}+\frac{3\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}=\frac{7\mathrm{\pi }}{6}$

(iv)
${\mathrm{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\mathrm{cot}}^{-1}\left(\frac{1}{\sqrt{3}}\right)+{\mathrm{tan}}^{-1}\left(\mathrm{sin}\left(-\frac{\mathrm{\pi }}{2}\right)\right)={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(-\frac{\mathrm{\pi }}{6}\right)\right]+{\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{3}\right)+{\mathrm{tan}}^{-1}\left(-1\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(-\frac{\mathrm{\pi }}{6}\right)\right]+{\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{3}\right)+{\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(-\frac{\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{12}$

#### Question 1:

(i) ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{6}\right)$
(ii) ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{7\mathrm{\pi }}{6}\right)$
(iii)  ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{5\mathrm{\pi }}{6}\right)$
(iv) ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{13\mathrm{\pi }}{7}\right)$
(v) ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{17\mathrm{\pi }}{8}\right)$
(vi) ${\mathrm{sin}}^{-1}\left\{\left(\mathrm{sin}-\frac{17\mathrm{\pi }}{8}\right)\right\}$
(vii) ${\mathrm{sin}}^{-1}\left(\mathrm{sin}3\right)$
(viii) ${\mathrm{sin}}^{-1}\left(\mathrm{sin}4\right)$
(ix) ${\mathrm{sin}}^{-1}\left(\mathrm{sin}12\right)$
(x) ${\mathrm{sin}}^{-1}\left(\mathrm{sin}2\right)$

#### Answer:

We know

(i) We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{6}\right)=\frac{\mathrm{\pi }}{6}$

(ii) We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{7\mathrm{\pi }}{6}\right)={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\mathrm{sin}-\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{6}$

(iii) We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{5\mathrm{\pi }}{6}\right)={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}$

(iv) We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{13\mathrm{\pi }}{7}\right)={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{7}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\mathrm{sin}-\frac{\mathrm{\pi }}{7}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{7}$

(v) We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{17\mathrm{\pi }}{8}\right)={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{8}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{8}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{8}$

(vi) We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}-\frac{17\mathrm{\pi }}{8}\right)={\mathrm{sin}}^{-1}\left(-\mathrm{sin}\frac{17\mathrm{\pi }}{8}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left\{-\mathrm{sin}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{8}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(-\mathrm{sin}\frac{\mathrm{\pi }}{8}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\mathrm{sin}-\frac{\mathrm{\pi }}{8}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{8}$

(vii) We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}3\right)={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\mathrm{\pi }-3\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }-3$

(viii)We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}4\right)={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\mathrm{\pi }-4\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }-4$

(ix) We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}12\right)={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(-\mathrm{\pi }+12\right)\right\}\phantom{\rule{0ex}{0ex}}=12-\mathrm{\pi }$

(x) )We have

${\mathrm{sin}}^{-1}\left(\mathrm{sin}2\right)={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\mathrm{\pi }-2\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }-2$

#### Question 2:

Evaluate each of the following:
(i)
${\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(-\frac{\mathrm{\pi }}{4}\right)\right\}$
(ii) ${\mathrm{cos}}^{-1}\left\{\mathrm{cos}\frac{5\mathrm{\pi }}{4}\right\}$
(iii)  ${\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{4\mathrm{\pi }}{3}\right)\right\}$
(iv) ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{13\mathrm{\pi }}{6}\right)$
(v) ${\mathrm{cos}}^{-1}\left(\mathrm{cos}3\right)$

(vi) ${\mathrm{cos}}^{-1}\left(\mathrm{cos}4\right)$

(vii) ${\mathrm{cos}}^{-1}\left(\mathrm{cos}5\right)$

(viii) ${\mathrm{cos}}^{-1}\left(\mathrm{cos}12\right)$

#### Answer:

We know

(i)  We have

${\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(-\frac{\mathrm{\pi }}{4}\right)\right\}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$

(ii) We have

${\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{5\mathrm{\pi }}{4}\right)\right\}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(2\mathrm{\pi }-\frac{3\mathrm{\pi }}{4}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{3\mathrm{\pi }}{4}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{3\mathrm{\pi }}{4}$

(iii) We have

${\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{4\mathrm{\pi }}{3}\right)\right\}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(2\mathrm{\pi }-\frac{2\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{2\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{\pi }}{3}$

(iv) We have

${\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{13\mathrm{\pi }}{6}\right)\right\}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(\frac{\mathrm{\pi }}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}$

(v) We have

${\mathrm{cos}}^{-1}\left(\mathrm{cos}3\right)=3$

(vi)We have

${\mathrm{cos}}^{-1}\left(\mathrm{cos}4\right)={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(2\mathrm{\pi }-4\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi }-4$

(vii) We have

${\mathrm{cos}}^{-1}\left(\mathrm{cos}5\right)={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(2\mathrm{\pi }-5\right)\right\}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi }-5$

(viii) We have

${\mathrm{cos}}^{-1}\left(\mathrm{cos}12\right)={\mathrm{cos}}^{-1}\left\{\mathrm{cos}\left(4\mathrm{\pi }-12\right)\right\}\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi }-12$

#### Question 3:

Evaluate each of the following:

(i) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{\mathrm{\pi }}{3}\right)$
(ii) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{6\mathrm{\pi }}{7}\right)$
(iii) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{7\mathrm{\pi }}{6}\right)$
(iv) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{9\mathrm{\pi }}{4}\right)$
(v) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}1\right)$
(v) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}2\right)$
(v) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}4\right)$
(v) ${\mathrm{tan}}^{-1}\left(\mathrm{tan}12\right)$

#### Answer:

We know that

(i) We have
${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{\mathrm{\pi }}{3}\right)=\frac{\mathrm{\pi }}{3}$
(ii) We have
${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{6\mathrm{\pi }}{7}\right)={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{7}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(-\frac{\mathrm{\pi }}{7}\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{7}$
(iii) We have

${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{7\mathrm{\pi }}{6}\right)={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}$
(iv) We have

${\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{9\mathrm{\pi }}{4}\right)={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$
(v) We have
${\mathrm{tan}}^{-1}\left(\mathrm{tan}1\right)=1$
(vi) We have

${\mathrm{tan}}^{-1}\left(\mathrm{tan}2\right)={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(-\mathrm{\pi }+2\right)\right]\phantom{\rule{0ex}{0ex}}=2-\mathrm{\pi }$

(vii) We have

${\mathrm{tan}}^{-1}\left(\mathrm{tan}4\right)={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(-\mathrm{\pi }+4\right)\right]\phantom{\rule{0ex}{0ex}}=4-\mathrm{\pi }$

(viii) We have

${\mathrm{tan}}^{-1}\left(\mathrm{tan}12\right)={\mathrm{tan}}^{-1}\left[\mathrm{tan}\left(-4\mathrm{\pi }+12\right)\right]\phantom{\rule{0ex}{0ex}}=12-4\mathrm{\pi }$

#### Question 4:

Evaluate each of the following:

(i) ${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{\mathrm{\pi }}{3}\right)$
(ii) ${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{2\mathrm{\pi }}{3}\right)$
(iii) ${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{5\mathrm{\pi }}{4}\right)$
(iv) ${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{7\mathrm{\pi }}{3}\right)$

(v) ${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{9\mathrm{\pi }}{5}\right)$
(vi)  ${\mathrm{sec}}^{-1}\left\{\mathrm{sec}\left(-\frac{7\mathrm{\pi }}{3}\right)\right\}$
(vii) ${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{13\mathrm{\pi }}{4}\right)$
(viii) ${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{25\mathrm{\pi }}{6}\right)$

#### Answer:

We know that

${\mathrm{sec}}^{1}\left(\mathrm{sec}\theta \right)=\theta$,      [0, π/2) â‹ƒ ( π/2,  π]

(i) We have
${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{\mathrm{\pi }}{3}\right)=\frac{\mathrm{\pi }}{3}$
(ii) We have
${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{2\mathrm{\pi }}{3}\right)=\frac{2\mathrm{\pi }}{3}$
(iii) We have

${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{5\mathrm{\pi }}{4}\right)={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(2\mathrm{\pi }-\frac{3\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(\frac{3\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{3\mathrm{\pi }}{4}$
(iv)We have

${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{7\mathrm{\pi }}{3}\right)={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}$

(v)We have

${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{9\mathrm{\pi }}{5}\right)={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{5}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(\frac{\mathrm{\pi }}{5}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{5}$
(vi) We have

${\mathrm{sec}}^{-1}\left\{\mathrm{sec}\left(-\frac{7\mathrm{\pi }}{3}\right)\right\}={\mathrm{sec}}^{-1}\left\{\mathrm{sec}\left(\frac{7\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}$
(vii)We have

${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{13\mathrm{\pi }}{4}\right)={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(4\mathrm{\pi }-\frac{3\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(\frac{3\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{3\mathrm{\pi }}{4}$
(viii)We have

${\mathrm{sec}}^{-1}\left(\mathrm{sec}\frac{25\mathrm{\pi }}{6}\right)={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(4\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{sec}}^{-1}\left[\mathrm{sec}\left(\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}$

#### Question 5:

Evaluate each of the following:

(i) ${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{\mathrm{\pi }}{4}\right)$
(ii) ${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{3\mathrm{\pi }}{4}\right)$
(iii) ${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{6\mathrm{\pi }}{5}\right)$
(iv) ${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{11\mathrm{\pi }}{6}\right)$
(v) ${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{13\mathrm{\pi }}{6}\right)$
(vi) ${\mathrm{cosec}}^{-1}\left\{\mathrm{cosec}\left(-\frac{9\mathrm{\pi }}{4}\right)\right\}$

#### Answer:

We know that

${\mathrm{cosec}}^{-1}\left(\mathrm{cosec\theta }\right)=\mathrm{\theta }$,      [−π/2, 0) â‹ƒ ( 0, π/2]

(i) ${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{\mathrm{\pi }}{4}\right)=\frac{\mathrm{\pi }}{4}$
(ii)
${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{3\mathrm{\pi }}{4}\right)={\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$
(iii)
${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{6\mathrm{\pi }}{5}\right)={\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{5}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cosec}}^{-1}\left(\mathrm{cosec}-\frac{\mathrm{\pi }}{5}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{5}$
(iv)
${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{11\mathrm{\pi }}{6}\right)={\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cosec}}^{-1}\left(\mathrm{cosec}-\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{6}$
(v)
${\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{13\mathrm{\pi }}{6}\right)={\mathrm{cosec}}^{-1}\left[\mathrm{cosec}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cosec}}^{-1}\left(\mathrm{cosec}\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}$
(vi)
${\mathrm{cosec}}^{-1}\left\{\mathrm{cosec}\left(-\frac{9\mathrm{\pi }}{4}\right)\right\}={\mathrm{cosec}}^{-1}\left[-\mathrm{cosec}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cosec}}^{-1}\left(-\mathrm{cosec}\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{cosec}}^{-1}\left(\mathrm{cosec}-\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{4}$

#### Question 6:

Evaluate each of the following:

(i) ${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{3}\right)$
(ii) ${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{4\mathrm{\pi }}{3}\right)$
(iii) ${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{9\mathrm{\pi }}{4}\right)$
(iv) ${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{19\mathrm{\pi }}{6}\right)$
(v) ${\mathrm{cot}}^{-1}\left\{\mathrm{cot}\left(-\frac{8\mathrm{\pi }}{3}\right)\right\}$
(vi) ${\mathrm{cot}}^{-1}\left\{\mathrm{cot}\left(\frac{21\mathrm{\pi }}{4}\right)\right\}$

#### Answer:

We know that

${\mathrm{cot}}^{-1}\left(\mathrm{cot\theta }\right)=\mathrm{\theta }$,      ( 0, π)

(i) We have
${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{3}\right)=\frac{\mathrm{\pi }}{3}$
(ii) We have
${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{4\mathrm{\pi }}{3}\right)={\mathrm{cot}}^{-1}\left[\mathrm{cot}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}$
(iii) We have
${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{9\mathrm{\pi }}{4}\right)={\mathrm{cot}}^{-1}\left[\mathrm{cot}\left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$
(iv) We have
${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{19\mathrm{\pi }}{6}\right)={\mathrm{cot}}^{-1}\left[\mathrm{cot}\left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}$
(v) We have
${\mathrm{cot}}^{-1}\left[\mathrm{cot}\left(-\frac{8\mathrm{\pi }}{3}\right)\right]={\mathrm{cot}}^{-1}\left[-\mathrm{cot}\left(\frac{8\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left[-\mathrm{cot}\left(3\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{3}$
(vi) We have
${\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{21\mathrm{\pi }}{4}\right)={\mathrm{cot}}^{-1}\left[\mathrm{cot}\left(5\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right)\right]\phantom{\rule{0ex}{0ex}}={\mathrm{cot}}^{-1}\left(\mathrm{cot}\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$

#### Question 7:

Write each of the following in the simplest form:

(i)

(ii)

(iii)

(iv)

(v)

(vi) ${\mathrm{tan}}^{-1}\sqrt{\frac{a-x}{a+x}},-a

(vii) ${\mathrm{tan}}^{-1}\left\{\frac{x}{a+\sqrt{{a}^{2}-{x}^{2}}}\right\},-a

(viii) ${\mathrm{sin}}^{-1}\left\{\frac{x+\sqrt{1-{x}^{2}}}{\sqrt{2}}\right\},-1

(ix)

(x)

#### Answer:

(i) Let $x=a\mathrm{sec}\theta$
Now,

(ii) Let $x=\mathrm{cot}\theta$
Now,

(iii) Let $x=\mathrm{cot}\theta$
Now,

${\mathrm{tan}}^{-1}\left\{\sqrt{1+{x}^{2}}-x\right\}={\mathrm{tan}}^{-1}\left\{\sqrt{1+{\mathrm{cot}}^{2}\theta }-\mathrm{cot}\theta \right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{\mathrm{cosec}\theta -\mathrm{cot}\theta \right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{\frac{1-\mathrm{cos}\theta }{\mathrm{sin}\theta }\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{\frac{2{\mathrm{sin}}^{2}\frac{\theta }{2}}{2\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2}}\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{\mathrm{tan}\left(\frac{\theta }{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{\theta }{2}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{cot}}^{-1}x}{2}\phantom{\rule{0ex}{0ex}}$

(iv) Let $x=\mathrm{tan}\theta$
Now,

(v) Let $x=\mathrm{tan}\theta$
Now,

(vi) Let $x=a\mathrm{cos}\theta$
Now,

(vii) Let $x=a\mathrm{sin}\theta$
Now,

(viii) Let $x=\mathrm{sin}\theta$
Now,

(ix) Let $x=\mathrm{cos}\theta$
Now,

(x) Let $x=\mathrm{cos}\theta$
Now,

#### Question 1:

Evaluate each of the following:

(i) $\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{7}{25}\right)$
(ii) $\mathrm{sin}\left({\mathrm{cos}}^{-1}\frac{5}{13}\right)$
(iii) $\mathrm{sin}\left({\mathrm{tan}}^{-1}\frac{24}{7}\right)$
(iv) $\mathrm{sin}\left({\mathrm{sec}}^{-1}\frac{17}{8}\right)$
(v) $\mathrm{cosec}\left({\mathrm{cos}}^{-1}\frac{3}{5}\right)$
(vi) $\mathrm{sec}\left({\mathrm{sin}}^{-1}\frac{12}{13}\right)$

(vii) $\mathrm{tan}\left({\mathrm{cos}}^{-1}\frac{8}{17}\right)$
(viii) $\mathrm{cot}\left({\mathrm{cos}}^{-1}\frac{3}{5}\right)$
(ix) $\mathrm{cos}\left({\mathrm{tan}}^{-1}\frac{24}{7}\right)$

#### Answer:

(i) $\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{7}{25}\right)=\frac{7}{25}$
(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

#### Question 2:

Prove the following results

(i) $\mathrm{tan}\left({\mathrm{cos}}^{-1}\frac{4}{5}+{\mathrm{tan}}^{-1}\frac{2}{3}\right)=\frac{17}{6}$
(ii) $\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{3}{5}+{\mathrm{cot}}^{-1}\frac{3}{2}\right)=\frac{6}{5\sqrt{13}}$

(iii) $\mathrm{tan}\left({\mathrm{sin}}^{-1}\frac{5}{13}+{\mathrm{cos}}^{-1}\frac{3}{5}\right)=\frac{63}{16}$

(iv) $\mathrm{sin}\left({\mathrm{cos}}^{-1}\frac{3}{5}+{\mathrm{sin}}^{-1}\frac{5}{13}\right)=\frac{63}{65}$

#### Answer:

(i)

(ii)
$\mathrm{LHS}=\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{3}{5}+{\mathrm{cot}}^{-1}\frac{3}{2}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{3}{5}+{\mathrm{tan}}^{-1}\frac{2}{3}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left[{\mathrm{cos}}^{-1}\sqrt{1-{\left(\frac{3}{5}\right)}^{2}}+{\mathrm{cos}}^{-1}\frac{1}{\sqrt{1+{\left(\frac{2}{3}\right)}^{2}}}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left({\mathrm{cos}}^{-1}\frac{4}{5}+{\mathrm{cos}}^{-1}\frac{3}{\sqrt{13}}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left\{{\mathrm{cos}}^{-1}\left[\frac{4}{5}×\frac{3}{\sqrt{13}}-\sqrt{1-{\left(\frac{4}{5}\right)}^{2}}\sqrt{1-{\left(\frac{3}{\sqrt{13}}\right)}^{2}}\right]\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left\{{\mathrm{cos}}^{-1}\left[\frac{12}{5\sqrt{13}}-\frac{6}{5\sqrt{13}}\right]\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left\{{\mathrm{cos}}^{-1}\left[\frac{6}{5\sqrt{13}}\right]\right\}\phantom{\rule{0ex}{0ex}}=\frac{6}{5\sqrt{13}}=\mathrm{RHS}$

(iii)
The question is wrong as we can't have arc sin greater than 1

(iv)
$\mathrm{LHS}=\mathrm{sin}\left({\mathrm{cos}}^{-1}\frac{3}{5}+{\mathrm{sin}}^{-1}\frac{5}{13}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left[{\mathrm{sin}}^{-1}\sqrt{1-{\left(\frac{3}{5}\right)}^{2}}+{\mathrm{sin}}^{-1}\frac{5}{13}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left[{\mathrm{sin}}^{-1}\frac{4}{5}+{\mathrm{sin}}^{-1}\frac{5}{13}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left\{{\mathrm{sin}}^{-1}\left[\frac{4}{5}×\sqrt{1-{\left(\frac{5}{13}\right)}^{2}}+\frac{5}{13}×\sqrt{1-{\left(\frac{4}{5}\right)}^{2}}\right]\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left[{\mathrm{sin}}^{-1}\left(\frac{48}{65}+\frac{15}{65}\right)\right]\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{63}{65}\right)\phantom{\rule{0ex}{0ex}}=\frac{63}{65}=\mathrm{RHS}$

#### Question 3:

Solve: $\mathrm{cos}\left({\mathrm{sin}}^{-1}x\right)=\frac{1}{6}$

#### Answer:

$\mathrm{cos}\left({\mathrm{sin}}^{-1}x\right)=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\left({\mathrm{cos}}^{-1}\sqrt{1-{x}^{2}}\right)=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\sqrt{1-{x}^{2}}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒1-{x}^{2}=\frac{1}{36}\phantom{\rule{0ex}{0ex}}⇒1-\frac{1}{36}={x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=\frac{35}{36}\phantom{\rule{0ex}{0ex}}⇒x=±\frac{\sqrt{35}}{6}$

#### Question 1:

Evaluate:
(i) $\mathrm{cos}\left\{{\mathrm{sin}}^{-1}\left(-\frac{7}{25}\right)\right\}$
(ii) $\mathrm{sec}\left\{{\mathrm{cot}}^{-1}\left(-\frac{5}{12}\right)\right\}$
(iii) $\mathrm{cot}\left\{{\mathrm{sec}}^{-1}\left(-\frac{13}{5}\right)\right\}$

#### Answer:

(i)
$\mathrm{cos}\left\{{\mathrm{sin}}^{-1}\left(-\frac{7}{25}\right)\right\}=\mathrm{cos}\left\{-{\mathrm{sin}}^{-1}\left(\frac{7}{25}\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left\{{\mathrm{sin}}^{-1}\left(\frac{7}{25}\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left\{{\mathrm{cos}}^{-1}\sqrt{1-{\left(\frac{7}{25}\right)}^{2}}\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left\{{\mathrm{cos}}^{-1}\frac{24}{25}\right\}\phantom{\rule{0ex}{0ex}}=\frac{24}{25}$
(ii)
$\mathrm{sec}\left\{{\mathrm{cot}}^{-1}\left(-\frac{5}{12}\right)\right\}=\mathrm{sec}\left\{\mathrm{\pi }-{\mathrm{cot}}^{-1}\left(\frac{5}{12}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{sec}\left\{{\mathrm{cot}}^{-1}\left(\frac{5}{12}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{sec}\left\{{\mathrm{cos}}^{-1}\left[\frac{1}{1+{\left(\frac{12}{5}\right)}^{2}}\right]\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{sec}\left\{{\mathrm{cos}}^{-1}\left(\frac{5}{13}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{sec}\left\{{\mathrm{sec}}^{-1}\left(\frac{13}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\frac{13}{5}$
(iii)
$\mathrm{cot}\left\{{\mathrm{sec}}^{-1}\left(-\frac{13}{5}\right)\right\}=\mathrm{cot}\left\{{\mathrm{sec}}^{-1}\left(\mathrm{\pi }-\frac{13}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{cot}\left\{{\mathrm{sec}}^{-1}\left(\frac{13}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{cot}\left\{{\mathrm{tan}}^{-1}\left(\frac{\sqrt{1-{\left(\frac{5}{13}\right)}^{3}}}{\frac{5}{13}}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{cot}\left\{{\mathrm{tan}}^{-1}\left(\frac{12}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{cot}\left\{{\mathrm{cot}}^{-1}\left(\frac{5}{12}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\frac{5}{12}$

#### Question 2:

Evaluate:
(i) $\mathrm{tan}\left\{{\mathrm{cos}}^{-1}\left(-\frac{7}{25}\right)\right\}$
(ii) $\mathrm{cosec}\left\{{\mathrm{cot}}^{-1}\left(-\frac{12}{5}\right)\right\}$
(iii) $\mathrm{cos}\left({\mathrm{tan}}^{-1}\frac{3}{4}\right)$

#### Answer:

(i)
$\mathrm{tan}\left\{{\mathrm{cos}}^{-1}\left(-\frac{7}{25}\right)\right\}=\mathrm{tan}\left\{{\mathrm{cos}}^{-1}\left(\mathrm{\pi }-\frac{7}{25}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{tan}\left\{{\mathrm{cos}}^{-1}\left(\frac{7}{25}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{tan}\left\{{\mathrm{tan}}^{-1}\left[\frac{\sqrt{1-{\left(\frac{7}{25}\right)}^{2}}}{\frac{7}{25}}\right]\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{tan}\left\{\mathrm{tan}\frac{24}{7}\right\}\phantom{\rule{0ex}{0ex}}=-\frac{24}{7}$
(ii)
$\mathrm{cosec}\left\{{\mathrm{cot}}^{-1}\left(-\frac{12}{5}\right)\right\}=\mathrm{cosec}\left\{{\mathrm{cot}}^{-1}\left(\mathrm{\pi }-\frac{12}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cosec}\left\{{\mathrm{cot}}^{-1}\left(\frac{12}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cosec}\left\{{\mathrm{sin}}^{-1}\left(\frac{\frac{5}{12}}{\sqrt{1+{\left(\frac{5}{12}\right)}^{2}}}\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cosec}\left\{{\mathrm{sin}}^{-1}\left(\frac{5}{13}\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{cosec}\left\{{\mathrm{cosec}}^{-1}\left(\frac{13}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{13}{5}$
(iii)  We have

Let $y={\mathrm{cos}}^{-1}\left(\frac{7}{25}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}y=\frac{7}{25}$
Now,

$\therefore \mathrm{cos}\left[{\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)\right]=\frac{4}{5}$

#### Question 3:

Evaluate: $\mathrm{sin}\left\{{\mathrm{cos}}^{-1}\left(-\frac{3}{5}\right)+{\mathrm{cot}}^{-1}\left(-\frac{5}{12}\right)\right\}$

#### Answer:

$\mathrm{sin}\left\{{\mathrm{cos}}^{-1}\left(-\frac{3}{5}\right)+{\mathrm{cot}}^{-1}\left(-\frac{5}{12}\right)\right\}=\mathrm{sin}\left\{\mathrm{\pi }-{\mathrm{cos}}^{-1}\left(\frac{3}{5}\right)+\mathrm{\pi }-{\mathrm{cot}}^{-1}\left(\frac{5}{12}\right)\right\}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\left\{2\mathrm{\pi }-\left[{\mathrm{cos}}^{-1}\left(\frac{3}{5}\right)+{\mathrm{cot}}^{-1}\left(\frac{5}{12}\right)\right]\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}\left\{{\mathrm{cos}}^{-1}\left(\frac{3}{5}\right)+{\mathrm{cot}}^{-1}\left(\frac{5}{12}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}\left\{{\mathrm{sin}}^{-1}\left[\sqrt{1-{\left(\frac{3}{5}\right)}^{2}}\right]+{\mathrm{sin}}^{-1}\left[\frac{\frac{12}{5}}{\sqrt{1+{\left(\frac{12}{5}\right)}^{2}}}\right]\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{4}{5}+{\mathrm{sin}}^{-1}\frac{12}{13}\right)\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}\left\{{\mathrm{sin}}^{-1}\left[\frac{4}{5}×\sqrt{1-{\left(\frac{12}{13}\right)}^{2}}+\frac{12}{13}×\sqrt{1-{\left(\frac{4}{5}\right)}^{2}}\right]\right\}\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}\left[{\mathrm{sin}}^{-1}\left(\frac{20}{65}+\frac{36}{65}\right)\right]\phantom{\rule{0ex}{0ex}}=-\mathrm{sin}\left[{\mathrm{sin}}^{-1}\left(\frac{56}{65}\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{56}{65}$

#### Question 1:

Find the principal value of each of the following:

(i) ${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
(ii) ${\mathrm{sin}}^{-1}\left(\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)$
(iii) ${\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)$
(iv) ${\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)$
(v) ${\mathrm{sin}}^{-1}\left(\mathrm{cos}\frac{3\mathrm{\pi }}{4}\right)$
(vi) ${\mathrm{sin}}^{-1}\left(\mathrm{tan}\frac{5\mathrm{\pi }}{4}\right)$

#### Answer:

(i) ${\mathrm{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(-\frac{\mathrm{\pi }}{3}\right)\right]=-\frac{\mathrm{\pi }}{3}$
(ii) ${\mathrm{sin}}^{-1}\left(\mathrm{cos}\frac{2\mathrm{\pi }}{3}\right)={\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(-\frac{\mathrm{\pi }}{6}\right)\right]=-\frac{\mathrm{\pi }}{6}$
(iii) ${\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)={\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{12}\right)=\frac{\mathrm{\pi }}{12}$
(iv) ${\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)={\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{5\mathrm{\pi }}{12}\right)=\frac{5\mathrm{\pi }}{12}$
(v) ${\mathrm{sin}}^{-1}\left(\mathrm{cos}\frac{3\mathrm{\pi }}{4}\right)={\mathrm{sin}}^{-1}\left(-\frac{\sqrt{2}}{2}\right)={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(-\frac{\mathrm{\pi }}{4}\right)\right]=-\frac{\mathrm{\pi }}{4}$
(vi) ${\mathrm{sin}}^{-1}\left(\mathrm{tan}\frac{5\mathrm{\pi }}{4}\right)=={\mathrm{sin}}^{-1}\left(1\right)={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}\right)\right]=\frac{\mathrm{\pi }}{2}$

#### Question 1:

Evaluate:

(i) $\mathrm{cot}\left({\mathrm{sin}}^{-1}\frac{3}{4}+{\mathrm{sec}}^{-1}\frac{4}{3}\right)$
(ii)
(iii)
(iv) $\mathrm{cot}\left({\mathrm{tan}}^{-1}a+{\mathrm{cot}}^{-1}a\right)$
(v)

(i)

(ii)

(iii)

(iv)

(v)

#### Question 2:

If ${\mathrm{cos}}^{-1}x+{\mathrm{cos}}^{-1}y=\frac{\mathrm{\pi }}{4}$, find the value of ${\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}y$

#### Question 3:

If ${\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}y=\frac{\mathrm{\pi }}{3}$ and ${\mathrm{cos}}^{-1}x-{\mathrm{cos}}^{-1}y=\frac{\mathrm{\pi }}{6}$, find the values of x and y.

#### Answer:

Solving ${\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}y=\frac{\mathrm{\pi }}{3}$ and ${\mathrm{sin}}^{-1}x-{\mathrm{sin}}^{-1}y=-\frac{\mathrm{\pi }}{6}$, we will get
$2{\mathrm{sin}}^{-1}x=\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}x=\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}⇒x=\mathrm{sin}\frac{\mathrm{\pi }}{12}=\frac{\sqrt{3}-1}{2\sqrt{2}}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}{\mathrm{sin}}^{-1}y=\frac{\mathrm{\pi }}{3}-{\mathrm{sin}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}y=\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}y=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒y=\mathrm{sin}\frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}$

#### Question 4:

If $\mathrm{cot}\left({\mathrm{cos}}^{-1}\frac{3}{5}+{\mathrm{sin}}^{-1}x\right)=0$, find the values of x.

#### Question 5:

If ${\left({\mathrm{sin}}^{-1}x\right)}^{2}+{\left({\mathrm{cos}}^{-1}x\right)}^{2}=\frac{17{\mathrm{\pi }}^{2}}{36}$, find x

#### Question 6:

$\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{1}{5}+{\mathrm{cos}}^{-1}x\right)=1$

#### Question 7:

${\mathrm{sin}}^{-1}x=\frac{\mathrm{\pi }}{6}+{\mathrm{cos}}^{-1}x$

#### Question 8:

$4{\mathrm{sin}}^{-1}x=\mathrm{\pi }-{\mathrm{cos}}^{-1}x$

#### Question 9:

${\mathrm{tan}}^{-1}x+2{\mathrm{cot}}^{-1}x=\frac{2\mathrm{\pi }}{3}$

#### Question 10:

$5{\mathrm{tan}}^{-1}x+3{\mathrm{cot}}^{-1}x=2\mathrm{\pi }$

#### Question 2:

(i) ${\mathrm{sin}}^{-1}\frac{1}{2}-2{\mathrm{sin}}^{-1}\frac{1}{\sqrt{2}}$
(ii) ${\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{\sqrt{3}}{2}\right)\right\}$

#### Answer:

(i)

${\mathrm{sin}}^{-1}\frac{1}{2}-2{\mathrm{sin}}^{-1}\frac{1}{\sqrt{2}}={\mathrm{sin}}^{-1}\frac{1}{2}-{\mathrm{sin}}^{-1}2×\frac{1}{\sqrt{2}}\sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^{2}}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\frac{1}{2}-{\mathrm{sin}}^{-1}\sqrt{2}×\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\frac{1}{2}-{\mathrm{sin}}^{-1}1\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{6}\right)-{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\mathrm{\pi }}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}=-\frac{\mathrm{\pi }}{3}$
(ii)
${\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{\sqrt{3}}{2}\right)\right\}={\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left({\mathrm{sin}}^{-1}\mathrm{sin}\frac{\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left\{\frac{1}{2}\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left\{\mathrm{sin}\frac{\mathrm{\pi }}{6}\right\}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}$

#### Question 3:

Find the domain of each of the following functions:

#### Answer:

(i)
To the domain of ${\mathrm{sin}}^{-1}y$ which is [−1, 1]
as x2 can not be negative

Hence, the domain is [−1, 1]

(ii)
Let f(x) = g(x) + h(x), where g(x)=cotx and h(x)=cot1x
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is [−1, 1]
The domain of h(x) is (−∞, ∞)
Therfore, the intersection of g(x) and h(x) is [−1, 1]
Hence, the domain is [−1, 1].

(iii)
To the domain of ${\mathrm{sin}}^{-1}y$ which is [−1, 1]
as square root can not be negative

Hence, the domain is

(iv)
Let f(x) = g(x) + h(x), where g(x)=cotx and h(x)=cot1x
Therefore, the domain of f(x) is given by the intersection of the domain of g(x) and h(x)
The domain of g(x) is [−1, 1]
The domain of h(x) is
Therfore, the intersection of g(x) and h(x) is
Hence, the domain is

#### Question 4:

If ${\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}y+{\mathrm{sin}}^{-1}z+{\mathrm{sin}}^{-1}t=2\mathrm{\pi }$, then find the value of x2 + y2 + z2 + t2

#### Answer:

We know that the maximum value of
Now,

Now,

#### Question 5:

If ${\left({\mathrm{sin}}^{-1}x\right)}^{2}+{\left({\mathrm{sin}}^{-1}y\right)}^{2}+{\left({\mathrm{sin}}^{-1}z\right)}^{2}=\frac{3}{4}{\mathrm{\pi }}^{2}$, find the value of x2 + y2 + z2

#### Answer:

We know that the maximum value of and minimum value of
Now,
For maximum value

and For minimum value

Now, For maximum value

and for minimum value

#### Question 1:

Prove the following results:

(i)
${\mathrm{tan}}^{-1}\frac{1}{7}+{\mathrm{tan}}^{-1}\frac{1}{13}={\mathrm{tan}}^{-1}\frac{2}{9}$
(ii) ${\mathrm{sin}}^{-1}\frac{12}{13}+{\mathrm{cos}}^{-1}\frac{4}{5}+{\mathrm{tan}}^{-1}\frac{63}{16}=\mathrm{\pi }$
(iii) ${\mathrm{tan}}^{-1}\frac{1}{4}+{\mathrm{tan}}^{-1}\frac{2}{9}={\mathrm{sin}}^{-1}\frac{1}{\sqrt{5}}$

(iii)

#### Question 2:

Find the value of ${\mathrm{tan}}^{-1}\left(\frac{x}{y}\right)-{\mathrm{tan}}^{-1}\left(\frac{x-y}{x+y}\right)$

#### Answer:

We know
${\mathrm{tan}}^{-1}x-{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\frac{x-y}{1+xy},xy>-1$
Now,
${\mathrm{tan}}^{-1}\left(\frac{x}{y}\right)-{\mathrm{tan}}^{-1}\left(\frac{x-y}{x+y}\right)$
$={\mathrm{tan}}^{-1}\left\{\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y}\left(\frac{x-y}{x+y}\right)}\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left\{\frac{\frac{{x}^{2}+xy-xy+{y}^{2}}{y\left(x+y\right)}}{\frac{{x}^{2}+{y}^{2}+xy-xy}{y\left(x+y\right)}}\right\}\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}1\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\mathrm{tan}\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{4}$

#### Question 3:

Solve the following equations for x:
(i) tan−12x + tan−13x = + $\frac{3\mathrm{\pi }}{4}$
(ii) tan−1(x + 1) + tan−1(x − 1) = tan−1$\frac{8}{31}$
(iii) tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x
(iv)
tan−1$\left(\frac{1-x}{1+x}\right)-\frac{1}{2}$tan−1x = 0, where x > 0
(v) cot
−1x − cot−1(x + 2) = $\frac{\mathrm{\pi }}{12}$> 0
(vi)  tan
−1(x + 2) + tan−1(x − 2) = tan−1$\left(\frac{8}{79}\right)$x > 0
(vii)

(viii) ${\mathrm{tan}}^{-1}\left(\frac{x-2}{x-4}\right)+{\mathrm{tan}}^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\mathrm{\pi }}{4}$
(ix)
(x) ${\mathrm{tan}}^{-1}\frac{x-2}{x-1}+{\mathrm{tan}}^{-1}\frac{x+2}{x+1}=\frac{\mathrm{\pi }}{4}$

#### Answer:

(i) We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

(ii)  We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

(iii) We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$ and ${\mathrm{tan}}^{-1}x-{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x-y}{1+xy}\right)$

(iv)

(v)

(vi) We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

(vii)  We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

(viii)
We know

${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

(ix)
We know
${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$

(x)

$⇒\frac{2{x}^{2}-4}{3}=1\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-4=3\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}=7\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=\frac{7}{2}\phantom{\rule{0ex}{0ex}}\therefore x=±\sqrt{\frac{7}{2}}$

#### Question 4:

Sum the following series:
${\mathrm{tan}}^{-1}\frac{1}{3}+{\mathrm{tan}}^{-1}\frac{2}{9}+{\mathrm{tan}}^{-1}\frac{4}{33}+...+{\mathrm{tan}}^{-1}\frac{{2}^{n-1}}{1+{2}^{2n-1}}$

#### Question 1:

Evaluate: $\mathrm{cos}\left({\mathrm{sin}}^{-1}\frac{3}{5}+{\mathrm{sin}}^{-1}\frac{5}{13}\right)$

#### Question 2:

(i) ${\mathrm{sin}}^{-1}\frac{63}{65}={\mathrm{sin}}^{-1}\frac{5}{13}+{\mathrm{cos}}^{-1}\frac{3}{5}$
(ii) ${\mathrm{sin}}^{-1}\frac{5}{13}+{\mathrm{cos}}^{-1}\frac{3}{5}={\mathrm{tan}}^{-1}\frac{63}{16}$
(iii) $\frac{9\mathrm{\pi }}{8}-\frac{9}{4}{\mathrm{sin}}^{-1}\frac{1}{3}=\frac{9}{4}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}$

(i)

(ii)

(iii)

#### Question 3:

Solve the following:

(i)  sin−1x + sin−12x = $\frac{\mathrm{\pi }}{3}$
(ii)  ${\mathrm{cos}}^{-1}x+{\mathrm{sin}}^{-1}\frac{x}{2}=\frac{\mathrm{\pi }}{6}$

#### Answer:

(i) We know
${\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}y={\mathrm{sin}}^{-1}\left[x\sqrt{1-{y}^{2}}+y\sqrt{1-{x}^{2}}\right]$

(ii)

#### Question 1:

If ${\mathrm{cos}}^{-1}\frac{x}{2}+{\mathrm{cos}}^{-1}\frac{y}{3}=\mathrm{\alpha },$ then prove that 9x2 − 12xy cos α + 4y2 = 36 sin2 α.

#### Answer:

We know
${\mathrm{cos}}^{-1}x+{\mathrm{cos}}^{-1}y={\mathrm{cos}}^{-1}\left[xy-\sqrt{1-{x}^{2}}\sqrt{1-{y}^{2}}\right]$
Now,

#### Question 2:

Solve the equation ${\mathrm{cos}}^{-1}\frac{a}{x}-{\mathrm{cos}}^{-1}\frac{b}{x}={\mathrm{cos}}^{-1}\frac{1}{b}-{\mathrm{cos}}^{-1}\frac{1}{a}$

#### Question 3:

Solve  ${\mathrm{cos}}^{-1}\sqrt{3}x+{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2}$

#### Answer:

$⇒3{x}^{4}=1-3{x}^{2}+3{x}^{4}-{x}^{2}\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒x=±\frac{1}{2}$

#### Question 4:

Prove that: ${\mathrm{cos}}^{-1}\frac{4}{5}+{\mathrm{cos}}^{-1}\frac{12}{13}={\mathrm{cos}}^{-1}\frac{33}{65}$

#### Answer:

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