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#### Question 1:

If sin $\mathrm{\theta }=\frac{\sqrt{3}}{2}$, find the value of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that sin $\theta$ = $\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$= $\frac{AB}{AC}$ = .

So, if AB = $\sqrt{3}k$, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2$-$ AB2 = (2k)2 $-$ ($\sqrt{3}k$)2
⇒ BC2 = 4k2$-$ 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos $\theta$  = $\frac{BC}{AC}$ =
tan $\theta$  =

∴ cot $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Question 2:

If cos  find the values of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and .
Now, we know that cos $\theta$ = = = .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2$-$ BC2 = (25k)2 $-$ (7k)2.
⇒ AB2 = 625k2$-$ 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin $\theta$ = =
tan $\theta$ =
∴ cot $\theta$ = , cosec $\theta$ = and sec $\theta$  =

#### Question 3:

If tan $\mathrm{\theta }=\frac{15}{8}$ find the values of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Base}}$ = $\frac{AB}{BC}$ = $\frac{15}{8}$.

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
sin $\theta$  = $\frac{AB}{AC}$ =
cos $\theta$  =

∴ cot $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Question 4:

If cot θ = 2, find the value of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cot $\theta$$\frac{\mathrm{base}}{\mathrm{Perpendicular}}$ = $\frac{BC}{AB}$ = 2.

So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = $\sqrt{5}$k
Now, finding the other T-ratios using their definitions, we get:
sin $\theta$  = $\frac{AB}{AC}$ =
cos $\theta$  =

∴ tan $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Question 5:

If cosec θ = $\sqrt{10}$, the find the values of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cosec $\theta$ = $\frac{\mathrm{Hypotenuse}}{\mathrm{Perpendicular}}$ = $\frac{AC}{AB}$= $\frac{\sqrt{10}}{1}$.

So, if AC = ($\sqrt{10}$)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2$-$ AB2 = 10k2$-$k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan $\theta$  = $\frac{AB}{BC}$ =

cos $\theta$  =

∴ , cot $\theta$  = and sec $\theta$  =

#### Question 6:

If $\mathrm{sin}\theta =\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}$, find the values of all T-ratios of $\theta$.

We have $\mathrm{sin}\theta =\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}$,

As,

${\mathrm{cos}}^{2}\theta =1-{\mathrm{sin}}^{2}\theta \phantom{\rule{0ex}{0ex}}=1-{\left(\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{1}-\frac{{\left({a}^{2}-{b}^{2}\right)}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left({a}^{2}+{b}^{2}\right)}^{2}-{\left({a}^{2}-{b}^{2}\right)}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left[\left({a}^{2}+{b}^{2}\right)-\left({a}^{2}-{b}^{2}\right)\right]\left[\left({a}^{2}+{b}^{2}\right)+\left({a}^{2}-{b}^{2}\right)\right]}{{\left({a}^{2}+{b}^{2}\right)}^{2}}$
$=\frac{\left[{a}^{2}+{b}^{2}-{a}^{2}+{b}^{2}\right]\left[{a}^{2}+{b}^{2}+{a}^{2}-{b}^{2}\right]}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left[2{b}^{2}\right]\left[2{a}^{2}\right]}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta =\frac{4{a}^{2}{b}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\sqrt{\frac{4{a}^{2}{b}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{2ab}{\left({a}^{2}+{b}^{2}\right)}$

Also,

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

Now,

$\mathrm{cosec}\theta =\frac{1}{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{2}+{b}^{2}}{{a}^{2}-{b}^{2}}$

Also,

$\mathrm{sec}\theta =\frac{1}{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{2ab}{{a}^{2}+{b}^{2}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{2}+{b}^{2}}{2ab}$

And,

$\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{{a}^{2}-{b}^{2}}{2ab}\right)}\phantom{\rule{0ex}{0ex}}=\frac{2ab}{{a}^{2}-{b}^{2}}$

#### Question 7:

If $15\mathrm{cot}A=8$, find the values of sinA and secA.

We have,

$15\mathrm{cot}A=8\phantom{\rule{0ex}{0ex}}⇒\mathrm{cot}A=\frac{8}{15}$

As,

${\mathrm{cosec}}^{2}A=1+{\mathrm{cot}}^{2}A\phantom{\rule{0ex}{0ex}}=1+{\left(\frac{8}{15}\right)}^{2}\phantom{\rule{0ex}{0ex}}=1+\frac{64}{225}\phantom{\rule{0ex}{0ex}}=\frac{225+64}{225}$
$⇒{\mathrm{cosec}}^{2}A=\frac{289}{225}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cosec}A=\sqrt{\frac{289}{225}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cosec}A=\frac{17}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sin}A}=\frac{17}{15}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}A=\frac{15}{17}$

Also,

${\mathrm{cos}}^{2}A=1-{\mathrm{sin}}^{2}A\phantom{\rule{0ex}{0ex}}=1-{\left(\frac{15}{17}\right)}^{2}\phantom{\rule{0ex}{0ex}}=1-\frac{225}{289}\phantom{\rule{0ex}{0ex}}=\frac{289-225}{289}$
$⇒{\mathrm{cos}}^{2}A=\frac{64}{289}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}A=\sqrt{\frac{64}{289}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}A=\frac{8}{17}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}A}=\frac{8}{17}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}A=\frac{17}{8}$

#### Question 8:

If $\mathrm{sin}A=\frac{9}{41}$, find the values of cosA and tanA.

We have $\mathrm{sin}A=\frac{9}{41}$,

As,

${\mathrm{cos}}^{2}A=1-{\mathrm{sin}}^{2}A\phantom{\rule{0ex}{0ex}}=1-{\left(\frac{9}{41}\right)}^{2}\phantom{\rule{0ex}{0ex}}=1-\frac{81}{1681}\phantom{\rule{0ex}{0ex}}=\frac{1681-81}{1681}$
$⇒{\mathrm{cos}}^{2}\mathrm{A}=\frac{1600}{1681}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}A=\sqrt{\frac{1600}{1681}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}A=\frac{40}{41}$

Also,

$\mathrm{tan}A=\frac{\mathrm{sin}A}{\mathrm{cos}A}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{9}{41}\right)}{\left(\frac{40}{41}\right)}\phantom{\rule{0ex}{0ex}}=\frac{9}{40}$

#### Question 9:

If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.

Let us consider a right $△$ABC right angled at B.
Now, we know that cos $\theta$ = 0.6 = $\frac{BC}{AC}$$\frac{3}{5}$

So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2  + BC2
⇒ AB2 = AC2$-$ BC2
⇒ AB2 = (5k)2$-$ (3k)2 = 25k2$-$ 9k2
⇒ AB2 = 16k2
⇒ AB = 4k

Finding out the other T-ratios using their definitions, we get:
sin $\theta$ =

tan $\theta$ =

Substituting the values in the given expression, we get:
5 sin $\theta$$-$ 3 tan $\theta$

i.e., LHS  = RHS

Hence proved.

#### Question 10:

If cosec θ = 2, show that $\left(\mathrm{cot\theta }+\frac{\mathrm{sin\theta }}{1+\mathrm{cos\theta }}\right)=2.$

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now, it is given that cosec $\theta$ = 2.
Also, sin $\theta$  =

So, if AB = k, then AC = 2k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC2 = AB2  + BC2
⇒ BC2 = AC2$-$ AB2
⇒ BC2 = (2k)2$-$ (k)2
⇒ BC2 = 3k2
⇒ BC = $\sqrt{3}$k
Finding out the other T-ratios using their definitions, we get:
cos $\theta$ =
tan $\theta$ =
cot $\theta$ =
Substituting these values in the given expression, we get:

$=\sqrt{3}+\frac{1}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}\left(2+\sqrt{3}\right)+1}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}+3+1}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\left(2+\sqrt{3}\right)}{2+\sqrt{3}}=2$
i.e., LHS = RHS

Hence proved.

#### Question 11:

If tan θ = $\frac{1}{\sqrt{7}}$, show that $\frac{\left({\mathrm{cosec}}^{2}\mathrm{\theta }-{\mathrm{sec}}^{2}\mathrm{\theta }\right)}{\left({\mathrm{cosec}}^{2}\mathrm{\theta }+{\mathrm{sec}}^{2}\mathrm{\theta }\right)}=\frac{3}{4}.$

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now it is given that tan $\theta$$\frac{AB}{BC}$$\frac{1}{\sqrt{7}}$.

So, if AB = k, then BC = $\sqrt{7}$k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2= AB2 + BC2
⇒ AC2 = (k)2 + ($\sqrt{7}$k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 2$\sqrt{2}$k
Now, finding out the values of the other trigonometric ratios, we have:
sin $\theta$  =
cos $\theta$  =
∴ cosec $\theta$  = and sec $\theta$   =
Substituting the values of cosec $\theta$  and sec $\theta$  in the given expression, we get:

i.e., LHS = RHS

Hence proved.

#### Question 12:

If tan θ = $\frac{20}{21}$, show that$\frac{\left(1-\mathrm{sin\theta }+\mathrm{cos\theta }\right)}{\left(1+\mathrm{sin\theta }+\mathrm{cos\theta }\right)}=\frac{3}{7}.$

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$$\frac{AB}{BC}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒  AC = 29k
Now, sin $\theta$ = and cos $\theta$ =

Substituting these values in the given expression, we get:

∴ LHS = RHS

Hence proved.

#### Question 13:

If $\mathrm{sec}\theta =\frac{5}{4}$, show that $\frac{\left(\mathrm{sin}\theta -2\mathrm{cos}\theta \right)}{\left(\mathrm{tan}\theta -\mathrm{cot}\theta \right)}=\frac{12}{7}$.

We have,

$\mathrm{sec}\theta =\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{cos}\theta }=\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{4}{5}$

Also,

${\mathrm{sin}}^{2}\theta =1-{\mathrm{cos}}^{2}\theta \phantom{\rule{0ex}{0ex}}=1-{\left(\frac{4}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}=1-\frac{16}{25}\phantom{\rule{0ex}{0ex}}=\frac{9}{25}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{3}{5}$

Now,

$=\frac{\frac{3}{5}×\frac{4}{5}\left(\frac{3}{5}-2×\frac{4}{5}\right)}{{\left(\frac{3}{5}\right)}^{2}-{\left(\frac{4}{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{12}{25}\left(\frac{3}{5}-\frac{8}{5}\right)}{\left(\frac{9}{25}-\frac{16}{25}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\frac{12}{25}×\left(\frac{-5}{5}\right)}{\left(\frac{-7}{25}\right)}\phantom{\rule{0ex}{0ex}}=\frac{12}{7}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Question 14:

If $\mathrm{cot}\theta =\frac{3}{4}$, show that $\sqrt{\frac{\mathrm{sec}\theta -\mathrm{cosec}\theta }{\mathrm{sec}\theta +\mathrm{cosec}\theta }}=\frac{1}{\sqrt{7}}$.

$=\sqrt{\frac{\left(\frac{1}{4}\right)}{\left(\frac{7}{4}\right)}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{7}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Question 15:

If $\mathrm{sin}\theta =\frac{3}{4}$, show that $\sqrt{\frac{{\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta }{{\mathrm{sec}}^{2}\theta -1}}=\frac{\sqrt{7}}{3}$.

$\mathrm{LHS}=\sqrt{\frac{{\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta }{{\mathrm{sec}}^{2}\theta -1}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{{\mathrm{tan}}^{2}\theta }}\phantom{\rule{0ex}{0ex}}=\sqrt{{\mathrm{cot}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\mathrm{cot}\theta \phantom{\rule{0ex}{0ex}}=\sqrt{{\mathrm{cosec}}^{2}\theta -1}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{1}{\mathrm{sin}\theta }\right)}^{2}-1}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{1}{\left(\frac{3}{4}\right)}\right)}^{2}-1}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{4}{3}\right)}^{2}-1}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{16}{9}-1}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{16-9}{9}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{7}{9}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}}{3}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Question 16:

If $\mathrm{sin}\theta =\frac{a}{b}$, show that $\left(\mathrm{sec}\theta +\mathrm{tan}\theta \right)=\sqrt{\frac{b+a}{b-a}}$.

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

#### Question 17:

If $\mathrm{cos}\theta =\frac{3}{5}$, show that $\frac{\left(\mathrm{sin}\theta -\mathrm{cot}\theta \right)}{2\mathrm{tan}\theta }=\frac{3}{160}$.

$\mathrm{LHS}=\frac{\left(\mathrm{sin}\theta -\mathrm{cot}\theta \right)}{2\mathrm{tan}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\left(\mathrm{sin}\theta -\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\right)}{2\left(\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{{\mathrm{sin}}^{2}\theta -\mathrm{cos}\theta }{\mathrm{sin}\theta }\right)}{\left(\frac{2\mathrm{sin}\theta }{\mathrm{cos}\theta }\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\theta \left({\mathrm{sin}}^{2}\theta -\mathrm{cos}\theta \right)}{2{\mathrm{sin}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\theta \left(1-{\mathrm{cos}}^{2}\theta -\mathrm{cos}\theta \right)}{2\left(1-{\mathrm{cos}}^{2}\theta \right)}$
$=\frac{\frac{3}{5}\left[1-{\left(\frac{3}{5}\right)}^{2}-\frac{3}{5}\right]}{2\left[1-{\left(\frac{3}{5}\right)}^{2}\right]}\phantom{\rule{0ex}{0ex}}=\frac{\frac{3}{5}\left(\frac{1}{1}-\frac{9}{25}-\frac{3}{5}\right)}{2\left(1-\frac{9}{25}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\frac{3}{5}\left(\frac{25-9-15}{25}\right)}{2\left(\frac{25-9}{25}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\frac{3}{5}\left(\frac{1}{25}\right)}{2\left(\frac{16}{25}\right)}$
$=\frac{3}{5×2×16}\phantom{\rule{0ex}{0ex}}=\frac{3}{160}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Question 18:

If tan θ = $\frac{4}{3}$, show that (sin θ + cos θ) = $\frac{7}{5}$.

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now, we know that  tan $\theta$$\frac{AB}{BC}$$\frac{4}{3}$.

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (4k)2 + (3k)2
⇒ AC2 = 16k2+ 9k2 = 25k2
⇒ AC = 5k
Finding out the values of sin $\theta$ and cos $\theta$ using their definitions, we have:
sin $\theta$ =
cos $\theta$ =
Substituting these values in the given expression, we get:
(sin $\theta$ + cos $\theta$) =
i.e., LHS = RHS

Hence proved.

#### Question 19:

If tan θ = $\frac{a}{b}$, show that

It is given that tan .

LHS =
Dividing the numerator and denominator by cos $\theta$, we get:

(âˆµ tan )
Now, substituting the value of tan $\theta$ in the above expression, we get:

i.e., LHS = RHS

Hence proved.

#### Question 20:

If 3 tan θ = 4, show that $\frac{4\mathrm{cos\theta }-\mathrm{sin\theta }}{2\mathrm{cos\theta }+\mathrm{sin\theta }}=\frac{4}{5}.$

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that tan $\theta$ = $\frac{AB}{BC}$$\frac{4}{3}$

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2  + BC2
⇒ AC2 = 16k2 + 9k2
⇒ AC2 = 25k2
⇒ AC = 5k

Now, we have:

sin $\theta$ =

cos $\theta$ =

Substituting these values in the given expression, we get:

i.e., LHS = RHS

Hence proved.

#### Question 21:

If 3 cot θ = 2, show that $\left(\frac{4\mathrm{sin\theta }-3\mathrm{cos\theta }}{2\mathrm{sin\theta }+6\mathrm{cos\theta }}\right)=\frac{1}{3}.$

It is given that cot .

LHS  =
Dividing the above expression by sin $\theta$, we get:
[âˆµ cot ]
Now, substituting the values of cot $\theta$ in the above expression, we get:

i.e., LHS = RHS

Hence proved.

#### Question 22:

If $3\mathrm{cot}\theta =4$, show that $\frac{\left(1-{\mathrm{tan}}^{2}\theta \right)}{\left(1+{\mathrm{tan}}^{2}\theta \right)}=\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)$.

$=\frac{\frac{16}{9}-1}{\frac{16}{9}+1}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{16-9}{9}\right)}{\left(\frac{16+9}{9}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{7}{9}\right)}{\left(\frac{25}{9}\right)}\phantom{\rule{0ex}{0ex}}=\frac{7}{25}$
$\mathrm{RHS}=\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)}{1}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }{{\mathrm{sin}}^{2}\theta }\right)}{\left(\frac{1}{{\mathrm{sin}}^{2}\theta }\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{{\mathrm{cos}}^{2}\theta }{{\mathrm{sin}}^{2}\theta }-\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{sin}}^{2}\theta }\right)}{{\mathrm{cosec}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\left({\mathrm{cot}}^{2}\theta -1\right)}{\left({\mathrm{cot}}^{2}\theta +1\right)}$
$=\frac{\left[{\left(\frac{4}{3}\right)}^{2}-1\right]}{\left[{\left(\frac{4}{3}\right)}^{2}+1\right]}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{16}{9}-\frac{1}{1}\right)}{\left(\frac{16}{9}+\frac{1}{1}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{16-9}{9}\right)}{\left(\frac{16+9}{9}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{7}{9}\right)}{\left(\frac{25}{9}\right)}\phantom{\rule{0ex}{0ex}}=\frac{7}{25}$

Since, LHS=RHS
Hence, verified.

#### Question 23:

If sec θ = $\frac{17}{8}$, verify that $\frac{3-4{\mathrm{sin}}^{2}\mathrm{\theta }}{4{\mathrm{cos}}^{2}\mathrm{\theta }-3}=\frac{3-{\mathrm{tan}}^{2}\mathrm{\theta }}{1-3{\mathrm{tan}}^{2}\mathrm{\theta }}$.

It is given that sec $\theta$ = $\frac{17}{8}$.

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that cos $\theta$ =

So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2$-$ BC2 = (17k)2$-$ (8k)2
⇒ AB2 = 289k2$-$ 64k2 = 225k2
⇒ AB = 15k.

Now, tan $\theta$  = and sin $\theta$ =

The given expression is .

Substituting the values in the above expression, we get:

∴ LHS = RHS
Hence proved.

#### Question 24:

In $∆$ABD,

Using Pythagoras theorem, we get

Again,

In $∆$ABC,

Using Pythagoras therem, we get

Now,

#### Question 25:

In a âˆ†ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.
Figure

Find the value of
(i) sin A
(ii) cos A
(iii) sin C
(iv) cos C

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (24)2 + (7)2
⇒ AC2 = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i)  sin A =

(ii) cos A =

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C =

(iv) cos C =

#### Question 26:

Given a âˆ†ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.
Show that (cos2θ − sin2θ) = $\frac{41}{841}.$

Using Pythagoras theorem, we get:
AB2  =  AC2  + BC2
⇒ AC2 = AB2$-$ BC2
⇒ AC2 = (29)2$-$ (21)2
⇒ AC2 = 841$-$ 441
⇒ AC2= 400
⇒ AC = $\sqrt{400}$ = 20 units

Now, sin and cos $\theta$ =

cos2$\theta$$-$ sin2$\theta$ =

Hence Proved.

#### Question 27:

In a âˆ†ABC, ∠B = 90°, AB = 12 cm and BC = 5 cm.

Find:
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = 122  + 52 = 144 + 25
⇒ AC2 = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, AB is base and BC is perpendicular
(i) cosA =
(ii) cosecA =

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cosC =
(iv) cosecC =

#### Question 28:

If $\mathrm{sin}\alpha =\frac{1}{2}$, prove that $\left(3\mathrm{cos}\alpha -4{\mathrm{cos}}^{3}\alpha \right)=0$.

$\mathrm{LHS}=\left(3\mathrm{cos}\alpha -4{\mathrm{cos}}^{3}\alpha \right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\alpha \left(3-4{\mathrm{cos}}^{2}\alpha \right)\phantom{\rule{0ex}{0ex}}=\sqrt{1-{\mathrm{sin}}^{2}\alpha }\left[3-4\left(1-{\mathrm{sin}}^{2}\alpha \right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{1-{\left(\frac{1}{2}\right)}^{2}}\left[3-4\left(1-{\left(\frac{1}{2}\right)}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{1}-\frac{1}{4}}\left[3-4\left(\frac{1}{1}-\frac{1}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{3}{4}}\left[3-4\left(\frac{3}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{3}{4}}\left[3-3\right]\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{3}{4}}\left[0\right]\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Question 29:

In a $∆$ABC, $\angle$B = 90$°$ and tanA = $\frac{1}{\sqrt{3}}$. Prove that
(i) sinA$·$cosC + cosA$·$sinC = 1            (ii) cosA$·$cosC $-$ sinA$·$sinC = 0

Now,

#### Question 30:

If $\angle$A and $\angle$B are acute angles such that sinA = sinB, then prove that $\angle$A = $\angle$B.

In $∆$ABC, $\angle$C = 90$°$
sinA = $\frac{\mathrm{BC}}{\mathrm{AB}}$ and
sinB = $\frac{\mathrm{AC}}{\mathrm{AB}}$

As, sinA = sinB
$⇒$$\frac{\mathrm{BC}}{\mathrm{AB}}$ = $\frac{\mathrm{AC}}{\mathrm{AB}}$
$⇒$BC = AC
So, $\angle$A = $\angle$B             (Angles opposite to equal sides are equal)

#### Question 31:

If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that tanA = tanB, the prove that $\angle \mathrm{A}=\angle \mathrm{B}$.

In ,

#### Question 32:

In a right $∆\mathrm{ABC}$, right-angled at $\mathrm{B}$, if $\mathrm{tanA}=1$, then verify that $2\mathrm{sinA}·\mathrm{cosA}=1$.

#### Question 33:

In the figure of . Find

In ,

Using Pythagoras theorem, we get

$\mathrm{PQ}=\sqrt{{\mathrm{PR}}^{2}-{\mathrm{QR}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(x+2\right)}^{2}-{x}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{x}^{2}+4x+4-{x}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4\left(x+1\right)}\phantom{\rule{0ex}{0ex}}=2\sqrt{x+1}$

Now,

#### Question 34:

If $x=\mathrm{cosecA}+\mathrm{cosA}$ and $y=\mathrm{cosecA}-\mathrm{cosA}$, then prove that ${\left(\frac{2}{x+y}\right)}^{2}+{\left(\frac{x-y}{2}\right)}^{2}-1=0$.

$\mathrm{LHS}={\left(\frac{2}{x+y}\right)}^{2}+{\left(\frac{x-y}{2}\right)}^{2}-1\phantom{\rule{0ex}{0ex}}={\left[\frac{2}{\left(\mathrm{cosecA}+\mathrm{cosA}\right)+\left(\mathrm{cosecA}-\mathrm{cosA}\right)}\right]}^{2}+{\left[\frac{\left(\mathrm{cosecA}+\mathrm{cosA}\right)-\left(\mathrm{cosecA}-\mathrm{cosA}\right)}{2}\right]}^{2}-1\phantom{\rule{0ex}{0ex}}={\left[\frac{2}{\mathrm{cosecA}+\mathrm{cosA}+\mathrm{cosecA}-\mathrm{cosA}}\right]}^{2}+{\left[\frac{\mathrm{cosecA}+\mathrm{cosA}-\mathrm{cosecA}+\mathrm{cosA}}{2}\right]}^{2}-1\phantom{\rule{0ex}{0ex}}={\left[\frac{2}{2\mathrm{cosecA}}\right]}^{2}+{\left[\frac{2\mathrm{cosA}}{2}\right]}^{2}-1$
$={\left[\frac{1}{\mathrm{cosecA}}\right]}^{2}+{\left[\mathrm{cosA}\right]}^{2}-1\phantom{\rule{0ex}{0ex}}={\left[\mathrm{sinA}\right]}^{2}+{\left[\mathrm{cosA}\right]}^{2}-1\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cos}}^{2}\mathrm{A}-1\phantom{\rule{0ex}{0ex}}=1-1\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Question 35:

If $x=\mathrm{cotA}+\mathrm{cosA}$ and $y=\mathrm{cotA}-\mathrm{cosA}$, prove that ${\left(\frac{x-y}{x+y}\right)}^{2}+{\left(\frac{x-y}{2}\right)}^{2}=1$.