# RD Sharma Solutions Class 10 Trigonometric Identities Exercise 6.2

## RD Sharma Solutions Class 10 Chapter 6 Exercise 6.2

### RD Sharma Class 10 Solutions Chapter 6 Ex 6.2 PDF Download

#### Exercise 6.2

Q1) If $cos\theta =\frac{4}{5}$, find all other trigonometric ratios of angle $\Theta$.

Solution:

We have:

$sin\Theta=\sqrt{1-cos^{2}\Theta}=\sqrt{1-(\frac{4}{5})^{2}}$

= $\sqrt{1-\frac{16}{25}}$

= $\sqrt{\frac{25-16}{25}}$

= $\sqrt{\frac{9}{25}}=\frac{3}{5}$

Therefore, $sin\Theta=\frac{3}{5}$

$tan\Theta=\frac{sin\Theta}{cos\Theta}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}sec\Theta=\frac{1}{cos\Theta}=\frac{1}{\frac{4}{5}}=\frac{5}{4}$

i.e. $cosec\Theta=\frac{1}{sec\Theta}=\frac{1}{\frac{3}{5}}=\frac{5}{3}cot\Theta=\frac{1}{tan\Theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

Q2) If $sin\Theta =\frac{1}{\sqrt{2}}$, find all other trigonometric ratios of angle $\Theta$.

Solution:

We have,

$cos\Theta=\sqrt{1-sin^{2}\Theta}=\sqrt{1-(\frac{1}{\sqrt{2}})^{2}}$

= $\sqrt{1-\frac{1}{2}}$

= $\sqrt{\frac{2-1}{2}}$

= $cos\Theta=\frac{1}{\sqrt{2}}$

= $tan\Theta=\frac{sin\Theta}{cos\Theta}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$

= $cosec\Theta=\frac{1}{sin\Theta}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}$

= $sec\Theta=\frac{1}{cos\Theta}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}$

= $cot\Theta=\frac{1}{tan\Theta}=\frac{1}{1}=1$

Q3) If $tan\Theta=\frac{1}{\sqrt{2}}$, find the value of $\frac{cosec^{2}\Theta-sec^{2}\Theta}{cosec^{2}\Theta+cot^{2}\Theta}$.

Solution:

We know that $sec\Theta =\sqrt{1+tan^{2}\Theta}$

=$\sqrt{1+(\frac{1}{\sqrt{2}})^{2}}$

= $\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

= $cot\Theta=\frac{1}{tan\Theta}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}$

= $cosec\Theta=\sqrt{1+cot^{2}\Theta}=\sqrt{1+2}=\sqrt{3}$

Substituting it in equation (1) we get

= $\frac{(\sqrt{3})^{2}-(\sqrt{\frac{3}{2}})^{2}}{(\sqrt{3})^{2}+(\sqrt{2})^{2}}=\frac{3-\frac{3}{2}}{3+2}=\frac{\frac{3}{2}}{5}=\frac{3}{10}$

Q4) If $tan\Theta=\frac{3}{4}$, find the value of $\frac{1-cos\Theta}{1+cos\Theta}$

Solution:

We know that

$sec\Theta =\sqrt{1+tan^{2}\Theta}$

= $\sqrt{1+(\frac{3}{4})^{2}}$

= $\sqrt{1+\frac{9}{16}}$

=$\sqrt{\frac{16+9}{16}}$

= $\sqrt{\frac{25}{16}}$

= $sec\Theta=\frac{5}{4}$

= $sec\Theta=\frac{1}{cos\Theta}=\frac{1}{\frac{5}{4}}=\frac{4}{5}=cos\Theta$

Therefore, $We\;get\;\frac{1-\frac{4}{5}}{1+\frac{4}{5}}=\frac{\frac{1}{5}}{\frac{9}{5}}=\frac{1}{9}$

Q5) If $tan\Theta=\frac{12}{5}$, find the value of $\frac{1+sin\Theta}{1-sin\Theta}$.

Solution:

$cot\Theta =\frac{1}{tan\Theta}=\frac{1}{\frac{12}{5}}=\frac{5}{12}$

= $cosec\Theta=\sqrt{1+cot^{2}\Theta}=\sqrt{1+[\frac{5}{12}]^{2}}=\sqrt{\frac{144+25}{144}}=\sqrt{\frac{169}{144}}=\frac{13}{12}$

= $sin\Theta=\frac{1}{cosec\Theta}=\frac{1}{\frac{13}{12}}=\frac{12}{13}$

i.e. $We\;get\;\frac{1+\frac{12}{13}}{1-\frac{12}{13}}=\frac{\frac{13+12}{18}}{\frac{13-12}{18}}=\frac{25}{1}=25$.

Q6) If $cot\Theta=\frac{1}{\sqrt{3}}$, find the value of $\frac{1-cos^{2}\Theta}{2-sin^{2}\Theta}$.

Solution:

$cosec\Theta =\sqrt{1+cot^{2}\Theta}=\sqrt{1+\frac{1}{3}}=\sqrt{\frac{4}{3}}$

= $cosec\Theta=\frac{2}{\sqrt{3}}$

= $sin\Theta=\frac{1}{cosec\Theta}=\frac{1}{\frac{2}{\sqrt{3}}}=\frac{\sqrt{3}}{2}$

= $and\;\frac{1}{cot\Theta}=\frac{sin\Theta}{cos\Theta}=cos\Theta=sin\Theta\times cot\Theta=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}=\frac{1}{2}$

Therefore, on substituting we get

= $\frac{1-(\frac{1}{2})^{2}}{2-(\frac{\sqrt{3}}{2})^{2}}=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{5}$.

Q7) If $cosecA=\sqrt{2}$, find the value of $\frac{2sin^{2}A+3cot^{2}A}{4(tan^{2}A-cos^{2}A)}$.

Solution:

We know that $cotA=\sqrt{cosec^{2}A-1}$

= $\sqrt{(2)^{2}-1}=\sqrt{2-1}$ =1.

= $tanA=\frac{1}{cotA}=\frac{1}{1}=1$

= $sinA=\frac{1}{cosecA}=\frac{1}{\sqrt{2}}$

= $sinA=\frac{1}{\sqrt{2}}$

$cosA=\sqrt{1-sin^{2}A}=\sqrt{1-(\frac{1}{\sqrt{2}})^{2}}=\sqrt{\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}}$

On substituting we get:

= $\frac{2[\frac{1}{\sqrt{2}}]^{2}+3[1]^{2}}{4[[1]-[\frac{1}{\sqrt{2}}]^{2}]}=\frac{2\times \frac{1}{2}+3}{4[1-\frac{1}{2}]}$

$\Rightarrow \frac{1+3}{4.\frac{1}{2}}=\frac{4}{2}=2$

Q8) If  $cot\Theta=\sqrt{3}$, find the value of $\frac{cosec^{2}\Theta+cot^{2}\Theta}{cosec^{2}\Theta-sec^{2}\Theta}$.

Solution:

$cosec\Theta=\sqrt{1+cot^{2}\Theta}=\sqrt{1+(\sqrt{3})^{2}}=\sqrt{1+3}=2$

$sin\Theta=\frac{1}{cosec\Theta}=\frac{1}{2}cot\Theta=\frac{cos\Theta}{sin\Theta}\;\;\;\;\;\;\cos\Theta=cot\Theta.sin\Theta$

$\Rightarrow cos\Theta=\frac{\sqrt{3}}{2}$

= $sec\Theta=\frac{1}{cos\Theta}=\frac{2}{\sqrt{3}}$

On substituting we get:

$\frac{(2)^{2}+(\sqrt{3})^{2}}{(2)^{2}-(\frac{2}{\sqrt{3}})^{2}}=\frac{4+3}{\frac{12-4}{3}}=\frac{7}{\frac{8}{3}}$

= $\frac{21}{8}$

Q9) If $3cos\Theta=1$, find the value of $\frac{6sin^{2}\Theta+tan^{2}\Theta}{4cos\Theta}$.

Solution:

$cos\Theta=\frac{1}{3},\;\;\;\;\;\;\;sin\Theta=\sqrt{1-cos^{2}\Theta}$

= $\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

$tan\Theta=\frac{sin\Theta}{cos\Theta}=\frac{2\sqrt{2}}{3.\frac{1}{3}}=2\sqrt{2}$

On substituting we get

$\frac{6[\frac{2\sqrt{2}}{3}]^{2}+(2\sqrt{2})^{2}}{4.\frac{1}{3}}=\frac{\frac{16}{3}+8}{\frac{4}{3}}=\frac{\frac{16+24}{3}}{\frac{4}{3}}$

= $\frac{40}{4}=10$

Q10) If $\sqrt{3}tan\Theta=sin\Theta$, find the value of $sin^{2}\Theta-cos^{2}\Theta$.

Solution:

$\sqrt{3}\frac{sin\Theta}{cos\Theta}=sin\Theta$

= $cos\Theta=\frac{\sqrt{3}}{3}\Rightarrow \frac{1}{\sqrt{3}}$

= $sin\Theta=\sqrt{1-cos^{2}\Theta}=\sqrt{1-(\frac{1}{\sqrt{3}})^{2}}$

= $sin^{2}\Theta-cos^{2}\Theta=(\sqrt{\frac{2}{3}})^{2}-(\frac{1}{\sqrt{3}})^{2}$

= $\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

Q11) If $cosec\Theta=\frac{13}{12}$, find the value of $\frac{2sin\Theta-3cos\Theta}{4sin\Theta-9cos\Theta}$.

Solution:

$sin\Theta=\frac{1}{cosec\Theta}=\frac{1}{\frac{13}{12}}=\frac{12}{13}$

= $cos\Theta=\sqrt{1-sin^{2}\Theta}=\sqrt{1-[\frac{12}{13}]^{2}}=\sqrt{1-\frac{144}{169}}$

= $\sqrt{\frac{25}{169}}=\frac{5}{13}$

$\Rightarrow \frac{2.\frac{12}{13}-3.\frac{5}{13}}{4.\frac{12}{13}-9.\frac{5}{13}}=\frac{\frac{24-15}{13}}{\frac{48-15}{13}}=\frac{9}{3}=3$

Q12) If $sin\Theta+cos\Theta=\sqrt{2}cos(90^{\circ}-\Theta)$, find $cot\Theta$.

Solution:

= $sin\Theta+cos\Theta=\sqrt{2}sin\Theta\;\;\;\;\;\;[cos(90-\Theta)=sin\Theta]$

$\Rightarrow cos\Theta=\sqrt{2}sin\Theta-sin\Theta\\ \Rightarrow cos\Theta=sin\Theta(\sqrt{2}-1)$

Divide both sides with $sin\Theta$ we get

= $\frac{cos\Theta}{sin\Theta}=\frac{sin\Theta}{sin\Theta}(\sqrt{2}-1)$

= $cot\Theta=\sqrt{2}-1$.

Q-13. If $2sin^{2}\Theta – cos^{2}\Theta = 2$, then find the value of $\Theta$.

Solution.

$2sin^{2}\Theta – cos^{2}\Theta = 2$

$\Rightarrow 2sin^{2}\Theta -\left ( 1 – sin^{2}\Theta \right ) = 2$

$\Rightarrow 2sin^{2}\Theta – 1 + sin^{2}\Theta = 2$

$\Rightarrow 3sin^{2}\Theta = 3$

$\Rightarrow sin^{2}\Theta = 1$

$\Rightarrow sin \Theta = 1$

$\Rightarrow sin\Theta = sin 90^{\circ}$

$\Rightarrow \Theta = 90^{\circ}$

Q-14. If $\sqrt{3} tan\Theta – 1 = 0$, find the value of $sin^{2}\Theta – cos^{2}\Theta$.

Solution.

$\sqrt{3} tan\Theta – 1 = 0$

$\Rightarrow \sqrt{3} tan\Theta = 1$

$\Rightarrow \sqrt{3} tan\Theta = \frac{1}{\sqrt{3}}$

$\sqrt{3} tan\Theta = tan 30^{\circ}$

$\Theta = 30^{\circ}$

Now,

$sin^{2}\Theta – cos^{2}\Theta$

= $sin^{2}\left ( 30^{\circ} \right ) – cos^{2}\left ( 30^{\circ} \right )$

= $\left ( \frac{1}{2} \right )^{2} – \left ( \frac{\sqrt{3}}{2} \right )^{2}$

= $\frac{1}{4} – \frac{3}{4}$ = $\frac{-2}{4}$ = $\frac{-1}{2}$