Rd Sharma Xi 2019 Solutions for Class 11 Science Math Chapter 9 Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle are provided here with simple step-by-step explanations. These solutions for Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle are extremely popular among Class 11 Science students for Math Values Of Trigonometric Functions At Multiples And Submultiples Of An Angle Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2019 Book of Class 11 Science Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2019 Solutions. All Rd Sharma Xi 2019 Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.

Prove that:

Prove that:

Prove that:

Prove that:

#### Question 5:

Prove that:

$\mathrm{LHS}=\frac{1-\mathrm{cos}2x+\mathrm{sin}2x}{1+\mathrm{cos}2x+\mathrm{sin}2x}$

Prove that:

#### Question 7:

Prove that:

$\mathrm{LHS}=\frac{\mathrm{cos}2x}{1+\mathrm{sin}2x}$

On dividing the numerator and denominator by cos x, we get

#### Question 8:

Prove that:

On dividing the numerator and denominator by $\mathrm{cos}\frac{x}{2}\phantom{\rule{0ex}{0ex}}$, we get

#### Question 9:

Prove that:

${\mathrm{cos}}^{2}\frac{\pi }{8}+{\mathrm{cos}}^{2}\frac{3\pi }{8}+{\mathrm{cos}}^{2}\frac{5\pi }{8}+{\mathrm{cos}}^{2}\frac{7\pi }{8}=2$

#### Question 10:

Prove that:

${\mathrm{sin}}^{2}\frac{\pi }{8}+{\mathrm{sin}}^{2}\frac{3\pi }{8}+{\mathrm{sin}}^{2}\frac{5\pi }{8}+{\mathrm{sin}}^{2}\frac{7\pi }{8}=2$

Prove that:

#### Question 12:

Prove that:

Using the identity $\mathrm{cos}C-\mathrm{cos}D=-2\mathrm{sin}\frac{C+D}{2}\mathrm{sin}\frac{C-D}{2}$, we get

#### Question 13:

Prove that:

$\mathrm{LHS}=1+{\mathrm{cos}}^{2}2x$
Using the identity $\mathrm{cos}2x={\mathrm{cos}}^{2}x-s{\mathrm{in}}^{2}x$, we get

#### Question 14:

Prove that:

Using the identity , we get

#### Question 15:

Prove that:

Using the identities , we get

Prove that:

Prove that:

#### Question 18:

Prove that:

Now, using the identities, we get

Show that:

Show that:

Prove that:

Prove that:

Prove that:

Prove that:

Prove that

#### Question 27:

Prove that:

$⇒{\mathrm{cot}}^{2}{\left(22\frac{1}{2}\right)}^{\circ }-2\mathrm{cot}{\left(22\frac{1}{2}\right)}^{\circ }-1=0\phantom{\rule{0ex}{0ex}}⇒\left\{{\mathrm{cot}}^{2}{\left(22\frac{1}{2}\right)}^{\circ }-2\mathrm{cot}{\left(22\frac{1}{2}\right)}^{\circ }+1\right\}-2=0\phantom{\rule{0ex}{0ex}}⇒{\left\{\mathrm{cot}{\left(22\frac{1}{2}\right)}^{\circ }-1\right\}}^{2}=2\phantom{\rule{0ex}{0ex}}⇒\mathrm{cot}{\left(22\frac{1}{2}\right)}^{\circ }-1=\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cot}{\left(22\frac{1}{2}\right)}^{\circ }=\sqrt{2}+1$

#### Question 28:

(i) If and x lies in the IIIrd quadrant, find the values of
.

(ii) If and x lies in IInd quadrant, find the values of sin 2x and $\mathrm{sin}\frac{x}{2}$

(i)

Using the identity $\mathrm{cos}2\mathrm{\theta }={\mathrm{cos}}^{2}\mathrm{\theta }-{\mathrm{sin}}^{2}\mathrm{\theta }$, we get

$\mathrm{cosx}={\mathrm{cos}}^{2}\frac{x}{2}-{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒-\frac{3}{5}=2{\mathrm{cos}}^{2}\frac{x}{2}-1\phantom{\rule{0ex}{0ex}}⇒1-\frac{3}{5}=2{\mathrm{cos}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}=2{\mathrm{cos}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{5}={\mathrm{cos}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\frac{x}{2}=±\sqrt{\frac{1}{5}}$

It is given that x lies in the third quadrant. This means that $\frac{x}{2}$ lies in the second quadrant.

$\therefore \mathrm{cos}\frac{x}{2}=-\frac{1}{\sqrt{5}}$
Again,

$\mathrm{cos}x={\mathrm{cos}}^{2}\frac{x}{2}-{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒-\frac{3}{5}={\left(-\frac{1}{\sqrt{5}}\right)}^{2}-{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒-\frac{3}{5}=\frac{1}{5}-{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒-\frac{1}{5}-\frac{3}{5}=-{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{4}{5}={\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\frac{x}{2}=±\frac{2}{\sqrt{5}}$

It is given that x lies in the third quadrant. This means that $\frac{x}{2}$ lies in the second quadrant.

$\therefore \mathrm{sin}\frac{x}{2}=\frac{2}{\sqrt{5}}$

Since x lies in the third quadrant, sinx is negative.

(ii)
$\mathrm{sin}x=\sqrt{1-{\mathrm{cos}}^{2}x}=\sqrt{1-\left(\frac{-3}{5}\right)}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}x=±\frac{4}{5}$
Here, x lies in the second quadrant.

$\therefore \mathrm{sin}x=\frac{4}{5}$

We know,

sin2x = 2sinx cosx

$\therefore \mathrm{sin}2x=2×\frac{4}{5}×\left(-\frac{3}{5}\right)=-\frac{24}{25}$
Now,
$\mathrm{cos}x=1-2{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{sin}}^{2}\frac{x}{2}=1-\left(-\frac{3}{5}\right)=\frac{8}{5}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\frac{x}{2}=\frac{4}{5}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\frac{x}{2}=±\frac{2}{\sqrt{5}}$

Since x lies in the second quadrant, $\frac{x}{2}$ lies in the first quadrant.
$\therefore \mathrm{sin}\frac{x}{2}=\frac{2}{\sqrt{5}}$

#### Question 29:

If and x lies in IInd quadrant, find the values of

Given:
$\mathrm{sin}x=\frac{\sqrt{5}}{3}$
Using the identity $\mathrm{cos}x=\sqrt{1-{\mathrm{sin}}^{2}x}$, we get

$\mathrm{cos}x=\sqrt{1-{\mathrm{sin}}^{2}x}=\sqrt{1-{\left(\frac{\sqrt{5}}{3}\right)}^{2}}=±\frac{2}{3}$

Since x lies in the 2nd quadrant, cosx is negative.

$\therefore \mathrm{cos}x=-\frac{2}{3}$

Now,

$\mathrm{cos}x=1-2{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒-\frac{2}{3}=1-2{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\frac{x}{2}=±\sqrt{\frac{5}{6}}$

Since x lies in the 2nd quadrant, $\frac{x}{2}$ lies in the 1st quadrant.

$\therefore \mathrm{sin}\frac{x}{2}=\sqrt{\frac{5}{6}}$

Again,

#### Question 30:

(i) If 0 ≤ xπ and x lies in the IInd quadrant such that . Find the values of
(ii) If and x is acute, find tan 2x
(iii) If and $0, find the value of sin 4x.

(i)

$\therefore \mathrm{sin}x=\sqrt{1-{\mathrm{cos}}^{2}x}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{1}{4}\right)}^{2}=1-{\mathrm{cos}}^{2}x\phantom{\rule{0ex}{0ex}}⇒\frac{1}{16}-1=-{\mathrm{cos}}^{2}x\phantom{\rule{0ex}{0ex}}⇒\frac{15}{16}={\mathrm{cos}}^{2}x\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}x=±\frac{\sqrt{15}}{4}\phantom{\rule{0ex}{0ex}}$

Since x lies in the 2nd quadrant, cos x is negative.

Thus,

$\mathrm{cos}x=-\frac{\sqrt{15}}{4}$

Now, using the identity $\mathrm{cos}x=2{\mathrm{cos}}^{2}\frac{x}{2}-1$, we get

$-\frac{\sqrt{15}}{4}=2{\mathrm{cos}}^{2}\frac{x}{2}-1\phantom{\rule{0ex}{0ex}}⇒-\frac{\sqrt{15}}{8}={\mathrm{cos}}^{2}\frac{x}{2}-\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\frac{x}{2}=\frac{4-\sqrt{15}}{8}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\frac{x}{2}=±\frac{4-\sqrt{15}}{8}$

Since x lies in the 2nd quadrant and $\frac{x}{2}$ lies in the 1st quadrant, $\mathrm{cos}\frac{x}{2}$ is positive.

$\therefore \mathrm{cos}\frac{x}{2}=\frac{4-\sqrt{15}}{8}$

Again,

$\mathrm{cos}x={\mathrm{cos}}^{2}\frac{x}{2}-{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒-\frac{\sqrt{15}}{4}={\left(\sqrt{\frac{4-\sqrt{15}}{8}}\right)}^{2}-{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒-\frac{\sqrt{15}}{4}=\frac{4-\sqrt{15}}{8}-{\mathrm{sin}}^{2}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\frac{x}{2}=\frac{4+\sqrt{15}}{8}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\frac{x}{2}=±\sqrt{\frac{4+\sqrt{15}}{8}}=\sqrt{\frac{4+\sqrt{15}}{8}}\phantom{\rule{0ex}{0ex}}$

Now,

(ii)

Now,

Hence, the value of tan 2x is $\frac{24}{7}$.

(iii) and $0.

Since x lies in the 1st quadrant, cos x is positive.

Thus,

Now,

Hence, the value of sin 4x is $-\frac{336}{625}$.

#### Question 31:

If $\mathrm{tan}x=\frac{b}{a}$, then find the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$.                            [NCERT]

Given: $\mathrm{tan}x=\frac{b}{a}$

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

#### Question 32:

If and , show that cos 2A = sin 4B

Given:
and

Using the identity $\mathrm{tan}2B=\frac{2\mathrm{tan}B}{1-{\mathrm{tan}}^{2}B}$, we get

$\mathrm{tan}2B=\frac{2×\frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4}$
Now, using the identities , we get

∴ cos 2A = sin 4B

#### Question 33:

Prove that:

On dividing and multiplying by $2\mathrm{sin}{7}^{\circ }$, we get

#### Question 34:

Prove that:

On dividing and multiplying by $2\mathrm{sin}\frac{2\mathrm{\pi }}{15}$, we get

#### Question 35:

Prove that: $\mathrm{cos}\frac{\mathrm{\pi }}{5}\mathrm{cos}\frac{2\mathrm{\pi }}{5}\mathrm{cos}\frac{4\mathrm{\pi }}{5}\mathrm{cos}\frac{8\mathrm{\pi }}{5}=\frac{-1}{16}$

#### Question 36:

Prove that:

On dividing and multiplying by $2\mathrm{sin}\frac{\mathrm{\pi }}{65}$, we get

If prove that

Given:

#### Question 38:

If , prove that

(i)

(ii)

The given equations are .

(i)

Now, using the identity $\mathrm{sin}C+\mathrm{sin}D=2\mathrm{sin}\frac{C+D}{2}\mathrm{cos}\frac{C-D}{2}$ for the LHS of, we get

On dividing (1) by (2), we get

We know,

$\mathrm{sin\theta }=\frac{2\mathrm{tan}\frac{\mathrm{\theta }}{2}}{1+{\mathrm{tan}}^{2}\frac{\mathrm{\theta }}{2}}$
$\therefore \mathrm{sin}\left(\mathrm{\alpha }+\mathrm{\beta }\right)=\frac{2\mathrm{tan}\left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)}{1+{\mathrm{tan}}^{2}\left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\left(\mathrm{\alpha }+\mathrm{\beta }\right)=\frac{2×\frac{a}{b}}{1+\frac{{a}^{2}}{{b}^{2}}}=\frac{2ab}{{a}^{2}+{b}^{2}}$

(ii)

If , prove that

#### Question 40:

If , prove that $\mathrm{tan}\frac{x}{2}=±\mathrm{tan}\frac{\alpha }{2}\mathrm{tan}\frac{\beta }{2}$

Given:

...(1)

#### Question 41:

If , prove that

Equation can be written as

$⇒\frac{{\mathrm{cos}}^{2}x×\mathrm{cos\alpha }}{{\mathrm{cos}}^{2}x×{\mathrm{cos}}^{2}\mathrm{\alpha }-\left(1-{\mathrm{cos}}^{2}x\right)×{\mathrm{sin}}^{2}\mathrm{\alpha }}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{cos}}^{2}x×\mathrm{cos\alpha }}{{\mathrm{cos}}^{2}x×{\mathrm{cos}}^{2}\mathrm{\alpha }-{\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}\mathrm{\alpha }}=1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}x×\mathrm{cos\alpha }={\mathrm{cos}}^{2}x×{\mathrm{cos}}^{2}\mathrm{\alpha }-{\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}\mathrm{\alpha }\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}x×\mathrm{cos\alpha }={\mathrm{cos}}^{2}x\left({\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\alpha }\right)-{\mathrm{sin}}^{2}\mathrm{\alpha }\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}x×\mathrm{cos\alpha }={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}\mathrm{\alpha }$

#### Question 42:

If and sin , prove that $\mathrm{cos}\frac{\alpha -\beta }{2}=±\frac{5}{24}$

Squaring and adding equations and , we get

Now,

#### Question 43:

If , prove that $\mathrm{cos}\frac{\alpha -\beta }{2}=\frac{8}{\sqrt{65}}$

Given:
.
Now,
$\mathrm{cos}\alpha =\sqrt{1-{\mathrm{sin}}^{2}\mathrm{\alpha }}=\sqrt{1-{\left(\frac{4}{5}\right)}^{2}}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}\mathrm{sin\beta }=\sqrt{1-{\mathrm{cos}}^{2}\mathrm{\alpha }}=\sqrt{1-{\left(\frac{5}{13}\right)}^{2}}=\frac{12}{13}$

Now,

#### Question 44:

If has α and β as its roots, then prove that

(i) $\mathrm{tan}\alpha +\mathrm{tan}\beta =\frac{2b}{a+c}$             [NCERT EXEMPLAR]

(ii)

(iii) $\mathrm{tan}\left(\alpha +\beta \right)=\frac{b}{a}$                  [NCERT EXEMPLAR]

Given:

This a quadratic equation in terms of ${\mathrm{tan}}^{2}x$.

It is given that α and β are the roots of the given equation, so tanα and tanβ are the roots of (1).

Since tanα and tanβ are the roots of the equation . Therefore,

(i)

(ii)

(iii)

#### Question 45:

If $\mathrm{cos}\alpha +\mathrm{cos}\beta =0=\mathrm{sin}\alpha +\mathrm{sin}\beta$, then prove that $\mathrm{cos}2\alpha +\mathrm{cos}2\beta =-2\mathrm{cos}\left(\alpha +\beta \right)$.                     [NCERT EXEMPLAR]

Given: $\mathrm{cos}\alpha +\mathrm{cos}\beta =0=\mathrm{sin}\alpha +\mathrm{sin}\beta$

$\therefore \mathrm{cos}\alpha +\mathrm{cos}\beta =0$

Squaring on both sides, we get

Also,

$\mathrm{sin}\alpha +\mathrm{sin}\beta =0$

Squaring on both sides, we get

Subtracting (2) from (1), we get

Prove that:

Prove that:

Prove that:

#### Question 10:

Prove that for all values of x

#### Question 11:

Prove that for all values of x

#### Question 1:

Prove that:
${\mathrm{sin}}^{2}\frac{2\pi }{5}-{\mathrm{sin}}^{2-}\frac{\pi }{3}=\frac{\sqrt{5}-1}{8}$

Prove that:

Prove that:

Prove that:

#### Question 5:

Prove that:
$\mathrm{cos}\frac{\pi }{15}\mathrm{cos}\frac{2\pi }{15}\mathrm{cos}\frac{4\pi }{15}\mathrm{cos}\frac{7\pi }{15}=\frac{1}{16}$

$\mathrm{LHS}=\mathrm{cos}\frac{\mathrm{\pi }}{15}\mathrm{cos}\frac{2\mathrm{\pi }}{15}\mathrm{cos}\frac{4\mathrm{\pi }}{15}\mathrm{cos}\frac{7\mathrm{\pi }}{15}$

$=\frac{2\mathrm{sin}\frac{2\mathrm{\pi }}{15}×\mathrm{cos}\frac{2\mathrm{\pi }}{15}}{2×2\mathrm{sin}\frac{\mathrm{\pi }}{15}}\mathrm{cos}\frac{4\mathrm{\pi }}{15}\mathrm{cos}\frac{7\mathrm{\pi }}{15}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{sin}\frac{4\mathrm{\pi }}{15}×\mathrm{cos}\frac{4\mathrm{\pi }}{15}}{2×2×2\mathrm{sin}\frac{\mathrm{\pi }}{15}}\mathrm{cos}\frac{7\mathrm{\pi }}{15}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\frac{8\mathrm{\pi }}{15}}{2×2×2\mathrm{sin}\frac{\mathrm{\pi }}{15}}\mathrm{cos}\frac{7\mathrm{\pi }}{15}$

$=\frac{\mathrm{sin\pi }+\mathrm{sin}\frac{\mathrm{\pi }}{15}}{16\mathrm{sin}\frac{\mathrm{\pi }}{15}}\phantom{\rule{0ex}{0ex}}=\frac{0+\mathrm{sin}\frac{\mathrm{\pi }}{15}}{16\mathrm{sin}\frac{\mathrm{\pi }}{15}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\frac{\mathrm{\pi }}{15}}{16\mathrm{sin}\frac{\mathrm{\pi }}{15}}\phantom{\rule{0ex}{0ex}}=\frac{1}{16}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$
Hence proved.

Prove that:

Prove that:

Prove that:

#### Question 9:

Prove that :
$\mathrm{sin}\frac{\mathrm{\pi }}{5}\mathrm{sin}\frac{2\mathrm{\pi }}{5}\mathrm{sin}\frac{3\mathrm{\pi }}{5}\mathrm{sin}\frac{4\mathrm{\pi }}{5}=\frac{5}{16}$

Thus, LHS = RHS

Hence, $\mathrm{sin}\frac{\mathrm{\pi }}{5}\mathrm{sin}\frac{2\mathrm{\pi }}{5}\mathrm{sin}\frac{3\mathrm{\pi }}{5}\mathrm{sin}\frac{4\mathrm{\pi }}{5}=\frac{5}{16}$.

Prove that:

#### Question 1:

If , then write the value of k.

#### Question 2:

If $\mathrm{tan}\frac{x}{2}=\frac{m}{n}$, then write the value of m sin x + n cos x.

Given:

$\mathrm{tan}\frac{x}{2}=\frac{m}{n}$

#### Question 3:

If $\frac{\pi }{2}, then write the value of .

#### Question 4:

If the write the value of in the simplest form.

We have,

#### Question 5:

If , then write the value of .

We have,

#### Question 6:

If $\pi , then write the value of .

#### Question 7:

In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.

#### Question 8:

Write the value of .

#### Question 9:

If $\frac{\pi }{4}, then write the value of .

#### Question 10:

Write the value of

Proceeding in the same way, we get

#### Question 11:

If $\mathrm{tan}A=\frac{1-\mathrm{cos}B}{\mathrm{sin}B}$, then find the value of tan2A.

Given:

$⇒2A=B\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}2A=\mathrm{tan}B$

Hence, the value of tan2A is tanB.

#### Question 12:

If $\mathrm{sin}x+\mathrm{cos}x=a$, then find the value of ${\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x$.

Given: $\mathrm{sin}x+\mathrm{cos}x=a$

Squaring on both sides, we get

Now,

Hence, the required value is $\frac{1}{4}\left[4-3{\left({a}^{2}-1\right)}^{2}\right]$.

#### Question 13:

If $\mathrm{sin}x+\mathrm{cos}x=a$, find the value of $\left|\mathrm{sin}x-\mathrm{cos}x\right|$.

Given: $\mathrm{sin}x+\mathrm{cos}x=a$

Now,

${\left(\mathrm{sin}x+\mathrm{cos}x\right)}^{2}+{\left(\mathrm{sin}x-\mathrm{cos}x\right)}^{2}={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x+2\mathrm{sin}x\mathrm{cos}x+{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x-2\mathrm{sin}x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒{\left(\mathrm{sin}x+\mathrm{cos}x\right)}^{2}+{\left(\mathrm{sin}x-\mathrm{cos}x\right)}^{2}=2\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)\phantom{\rule{0ex}{0ex}}⇒{\left(\mathrm{sin}x+\mathrm{cos}x\right)}^{2}+{\left(\mathrm{sin}x-\mathrm{cos}x\right)}^{2}=2$

Thus, the required value is $\sqrt{2-{a}^{2}}$.

is equal to

(a) 8 cos x
(b) cos x
(c) 8 sin x
(d) sin x

(d) sin x

#### Question 2:

(a)

(b)

(c)

(d) none of these.

(b)

#### Question 3:

The value of is
(a) $\frac{1}{8}$

(b) $\frac{1}{16}$

(c) $\frac{1}{32}$

(d) none of these

(d) none of these

#### Question 4:

If then, is equal to
(a) 1

(b) $-1$

(c) $-\sqrt{5}$

(d) $\sqrt{5}$

(a) 1

#### Question 5:

For all real values of x, is equal to
(a)
(b)
(c)
(d) none of these

(b)

#### Question 6:

The value of is
(a) 0
(b) $\sqrt{5}$
(c) 1
(d) none of these

(a) 0

#### Question 7:

If in a , then
(a) 6
(b) 1
(c) $\frac{1}{6}$
(d) none of these

(d) none of these

ABC is a triangle.

#### Question 8:

If and , then $\lambda =$
(a) $\frac{1}{4}$

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

(c) 1

(d) none of these

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

#### Question 9:

If

(a)

(b)

(c)

(d) none of these

(a)

#### Question 10:

If , then

(a)

(b)

(c)

(d) none of these

(b)

#### Question 11:

If
(a) $-\frac{a}{b}$

(b) $-\frac{b}{a}$

(c) $\sqrt{{a}^{2}+{b}^{2}}$

(d) none of these

(b) $-\frac{b}{a}$

The value of is
(a) 1
(b) 2
(c) 3
(d) 4

(d) 4

#### Question 13:

The value of
(a) 1
(b) $-1$
(c)
(d) none of these.

(d) none of these

#### Question 14:

(a) 1
(b) 2
(c) 4
(d) none of these.

(b) 2

#### Question 15:

If , then is equal to
(a)
(b)
(c)
(d)

(c)

The value of is
(a) 2
(b) 1
(c) 0
(d) −1

(c) 0

#### Question 17:

If , then A lies in the interval
(a)
(b)
(c)
(d) none of these

(a)

#### Question 18:

The value of is equal to
(a) cos x
(b) sin x
(c) tan x
(d) none of these

(a) cos x

If
(a) 3
(b) 4
(c) 1
(d) 2

(d) 2

#### Question 20:

The value of is
(a)

(b) 0

(c)

(d) $\frac{1}{2}$

(a)

is equal to

(a) cos x
(b) sin x
(c) – cos x
(d) sin x

(b) sin x

#### Question 22:

The value of is
(a) 0
(b) cos 3A
(c) cos 2A
(d) none of these

(c) cos 2A

The value of is
(a) cos x
(b) sec x
(c) cosec x
(d) sin x

(c) cosec x

is equal to
(a)
(b)
(c)
(d)

(a)

#### Question 25:

If α and β are acute angles satisfying , then tan α =
(a)

(b)

(c)

(d)

(a)

#### Question 26:

If , then

(a)

(b)

(c)

(d)

(d)

$⇒\frac{{\mathrm{sin}}^{2}\frac{\mathrm{\alpha }}{2}}{{\mathrm{cos}}^{2}\frac{\mathrm{\alpha }}{2}}\left(1-\mathrm{e}\right)=\frac{{\mathrm{sin}}^{2}\frac{x}{2}}{{\mathrm{cos}}^{2}\frac{x}{2}}\left(1+\mathrm{e}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{1}{2}\left(1-\mathrm{cos\alpha }\right)}{\frac{1}{2}\left(1+\mathrm{cos\alpha }\right)}\left(1-\mathrm{e}\right)=\frac{\frac{1}{2}\left(1-\mathrm{cos}x\right)}{\frac{1}{2}\left(1+\mathrm{cos}x\right)}\left(1+\mathrm{e}\right)\phantom{\rule{0ex}{0ex}}⇒\left(1-\mathrm{cos\alpha }\right)\left(1+\mathrm{cos}x\right)\left(1-\mathrm{e}\right)=\left(1+\mathrm{cos\alpha }\right)\left(1-\mathrm{cos}x\right)\left(1+\mathrm{e}\right)\phantom{\rule{0ex}{0ex}}⇒\left(1+\mathrm{cos}x\right)\left(1-\mathrm{e}\right)-\mathrm{cos\alpha }\left(1+\mathrm{cos}x\right)\left(1-\mathrm{e}\right)=\left(1-\mathrm{cos}x\right)\left(1+\mathrm{e}\right)+\mathrm{cos\alpha }\left(1-\mathrm{cos}x\right)\left(1+\mathrm{e}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos\alpha }\left\{\left(1+\mathrm{cos}x\right)\left(1-\mathrm{e}\right)+\left(1-\mathrm{cos}x\right)\left(1+\mathrm{e}\right)\right\}=\left(1+\mathrm{cos}x\right)\left(1-\mathrm{e}\right)-\left(1-\mathrm{cos}x\right)\left(1+\mathrm{e}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos\alpha }=\frac{2\mathrm{cos}x-2\mathrm{e}}{2-2\mathrm{ecos}x}=\frac{\mathrm{cos}x-\mathrm{e}}{1-\mathrm{ecos}x}$

#### Question 27:

If then
(a) $-1$
(b) 1
(c) 1/2
(d) None of these

(b) 1

#### Question 28:

If then

(a) $\frac{1+t}{1-t}$

(b) $\frac{1-t}{1+t}$

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

(d) $\frac{2t}{1+t}$

(a) $\frac{1+t}{1-t}$

#### Question 29:

The value of is
(a) cos 2x
(b) sin 2x
(c) cos 4x
(d) none of these

(c) cos 4x

The value of  is
(a) cos 2A
(b) sin 2A
(c) cos A
(d) 0

(a) cos 2A

The value of is
(a) cot 3x
(b) 2cot 3x
(c) tan 3x
(d) 3 tan 3x

(c) tan 3x

The value of is
(a) 3 tan 3x
(b) tan 3x
(c) 3 cot 3x
(d) cot 3x

(a) 3 tan 3x

#### Question 33:

The value of
(a)
(b)
(c)
(d) None of these

(c)

is equal to
(a)
(b)
(c)
(d)

(a)

#### Question 35:

If is equal to

(a)

(b)

(c)

(d)

(e) None of these

(d)

#### Question 36:

If $\mathrm{tan}x=\frac{a}{b}$, then is equal to

(a) a

(b) b

(c) $\frac{a}{b}$

(d) $\frac{b}{a}$

Given: $\mathrm{tan}x=\frac{a}{b}$
Now,

$=\frac{{b}^{3}-{a}^{2}b+2{a}^{2}b}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{b}^{3}+{a}^{2}b}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{b\left({b}^{2}+{a}^{2}\right)}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}=b$

Hence, the correct answer is option B.
Given: $\mathrm{tan}x=\frac{a}{b}$
Now,

$=\frac{{b}^{3}-{a}^{2}b+2{a}^{2}b}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{b}^{3}+{a}^{2}b}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{b\left({b}^{2}+{a}^{2}\right)}{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}=b$

Hence, the correct answer is option B.

#### Question 37:

If $\mathrm{tan}\alpha =\frac{1}{7},\mathrm{tan}\beta =\frac{1}{3}$, then $\mathrm{cos}2\alpha$ is equal to

(a) $\mathrm{sin}2\beta$                                 (b) $\mathrm{sin}4\beta$                                 (c) $\mathrm{sin}3\beta$                                 (d) $\mathrm{cos}2\beta$

It is given that $\mathrm{tan}\alpha =\frac{1}{7}$ and $\mathrm{tan}\beta =\frac{1}{3}$.

Now,

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

$\therefore \mathrm{cos}2\alpha =\mathrm{sin}4\beta$

Hence, the correct answer is option B.

#### Question 38:

The value of ${\mathrm{cos}}^{2}48°-{\mathrm{sin}}^{2}12°$ is

(a) $\frac{\sqrt{5}+1}{8}$                             (b) $\frac{\sqrt{5}-1}{8}$                             (c) $\frac{\sqrt{5}+1}{5}$                             (d) $\frac{\sqrt{5}+1}{2\sqrt{2}}$