Class 10: Trigonometric Identities

$\displaystyle \textbf{Question 1: }\text{If }\sin\theta=\frac{1}{\sqrt{2}},\text{ find all other trigonometric ratios of angle }\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin\theta=\frac{1}{\sqrt{2}}$
$\displaystyle \cos\theta=\sqrt{1-\sin^2\theta}$
$\displaystyle =\sqrt{1-\frac{1}{2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$
$\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{1/\sqrt{2}}{1/\sqrt{2}}=1$
$\displaystyle \cot\theta=\frac{1}{\tan\theta}=1$
$\displaystyle \sec\theta=\frac{1}{\cos\theta}=\sqrt{2}$
$\displaystyle \mathrm{cosec}\theta=\frac{1}{\sin\theta}=\sqrt{2}$
$\displaystyle \therefore \cos\theta=\frac{1}{\sqrt{2}},\ \tan\theta=1,\ \cot\theta=1,\ \sec\theta=\sqrt{2},\ \mathrm{cosec}\theta=\sqrt{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{If }\tan\theta=\frac{1}{\sqrt{2}},\text{ find the value of } \frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\cot^2\theta}.$
$\displaystyle \text{Answer:}$
$\displaystyle \tan\theta=\frac{1}{\sqrt{2}}$
$\displaystyle \therefore \tan^2\theta=\frac{1}{2}$
$\displaystyle \sec^2\theta=1+\tan^2\theta=1+\frac{1}{2}=\frac{3}{2}$
$\displaystyle \cot\theta=\frac{1}{\tan\theta}=\sqrt{2}$
$\displaystyle \therefore \cot^2\theta=2$
$\displaystyle \mathrm{cosec}^2\theta=1+\cot^2\theta=1+2=3$
$\displaystyle \therefore \frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\cot^2\theta}$
$\displaystyle =\frac{3-\frac{3}{2}}{3+2}$
$\displaystyle =\frac{\frac{3}{2}}{5}=\frac{3}{10}$
$\displaystyle \therefore \text{The required value is }\frac{3}{10}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{If }4\tan\theta=3,\text{ evaluate }\frac{4\sin\theta-\cos\theta+1}{4\sin\theta+\cos\theta-1}.\hfill\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle 4\tan\theta=3$
$\displaystyle \therefore \tan\theta=\frac{3}{4}$
$\displaystyle \sec\theta=\sqrt{1+\tan^2\theta}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
$\displaystyle \therefore \cos\theta=\frac{1}{\sec\theta}=\frac{4}{5}$
$\displaystyle \sin\theta=\tan\theta\cos\theta=\frac{3}{4}\times\frac{4}{5}=\frac{3}{5}$
$\displaystyle \therefore \frac{4\sin\theta-\cos\theta+1}{4\sin\theta+\cos\theta-1}$
$\displaystyle =\frac{4\left(\frac{3}{5}\right)-\frac{4}{5}+1}{4\left(\frac{3}{5}\right)+\frac{4}{5}-1}$
$\displaystyle =\frac{\frac{12}{5}-\frac{4}{5}+1}{\frac{12}{5}+\frac{4}{5}-1}$
$\displaystyle =\frac{\frac{13}{5}}{\frac{11}{5}}=\frac{13}{11}$
$\displaystyle \therefore \text{The required value is }\frac{13}{11}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }\tan\theta=\frac{12}{5},\text{ find the value of }\frac{1+\sin\theta}{1-\sin\theta}.$
$\displaystyle \text{Answer:}$
$\displaystyle \tan\theta=\frac{12}{5}$
$\displaystyle \sec\theta=\sqrt{1+\tan^2\theta}=\sqrt{1+\frac{144}{25}}=\frac{13}{5}$
$\displaystyle \therefore \cos\theta=\frac{1}{\sec\theta}=\frac{5}{13}$
$\displaystyle \sin\theta=\tan\theta\cos\theta=\frac{12}{5}\times\frac{5}{13}=\frac{12}{13}$
$\displaystyle \therefore \frac{1+\sin\theta}{1-\sin\theta}=\frac{1+\frac{12}{13}}{1-\frac{12}{13}}$
$\displaystyle =\frac{\frac{25}{13}}{\frac{1}{13}}=25$
$\displaystyle \therefore \text{The required value is }25.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }\cot\theta=\frac{1}{\sqrt{3}},\text{ find the value of }\frac{1-\cos^2\theta}{2-\sin^2\theta}.$
$\displaystyle \text{Answer:}$
$\displaystyle \cot\theta=\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \tan\theta=\sqrt{3}$
$\displaystyle \sec\theta=\sqrt{1+\tan^2\theta}=\sqrt{1+3}=2$
$\displaystyle \therefore \cos\theta=\frac{1}{\sec\theta}=\frac{1}{2}$
$\displaystyle \sin\theta=\tan\theta\cos\theta=\sqrt{3}\times\frac{1}{2}=\frac{\sqrt{3}}{2}$
$\displaystyle \therefore \frac{1-\cos^2\theta}{2-\sin^2\theta}$
$\displaystyle =\frac{1-\frac{1}{4}}{2-\frac{3}{4}}$
$\displaystyle =\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{5}$
$\displaystyle \therefore \text{The required value is }\frac{3}{5}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }\mathrm{cosec} A=\sqrt{2},\text{ find the value of } \frac{2\sin^2 A+3\cot^2 A}{4(\tan^2 A-\cos^2 A)}.$
$\displaystyle \text{Answer:}$
$\displaystyle \mathrm{cosec} A=\sqrt{2}$
$\displaystyle \therefore \sin A=\frac{1}{\sqrt{2}}$
$\displaystyle \cos A=\sqrt{1-\sin^2 A}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$
$\displaystyle \therefore \tan A=\frac{\sin A}{\cos A}=1,\qquad \cot A=1$
$\displaystyle \therefore \frac{2\sin^2 A+3\cot^2 A}{4(\tan^2 A-\cos^2 A)}$
$\displaystyle =\frac{2\left(\frac{1}{2}\right)+3(1)^2}{4\left(1-\frac{1}{2}\right)}$
$\displaystyle =\frac{1+3}{2}$
$\displaystyle =2$
$\displaystyle \therefore \text{The required value is }2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }\cot\theta=\sqrt{3},\text{ find the value of } \frac{\mathrm{cosec}^2\theta+\cot^2\theta}{\mathrm{cosec}^2\theta-\sec^2\theta}.$
$\displaystyle \text{Answer:}$
$\displaystyle \cot\theta=\sqrt{3}$
$\displaystyle \therefore \cot^2\theta=3$
$\displaystyle \mathrm{cosec}^2\theta=1+\cot^2\theta=1+3=4$
$\displaystyle \tan\theta=\frac{1}{\cot\theta}=\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \tan^2\theta=\frac{1}{3}$
$\displaystyle \sec^2\theta=1+\tan^2\theta=1+\frac{1}{3}=\frac{4}{3}$
$\displaystyle \therefore \frac{\mathrm{cosec}^2\theta+\cot^2\theta}{\mathrm{cosec}^2\theta-\sec^2\theta}$
$\displaystyle =\frac{4+3}{4-\frac{4}{3}}$
$\displaystyle =\frac{7}{\frac{8}{3}}=\frac{21}{8}$
$\displaystyle \therefore \text{The required value is }\frac{21}{8}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }3\cos\theta=1,\text{ find the value of }\frac{6\sin^2\theta+\tan^2\theta}{4\cos\theta}.$
$\displaystyle \text{Answer:}$
$\displaystyle 3\cos\theta=1$
$\displaystyle \therefore \cos\theta=\frac{1}{3}$
$\displaystyle \sin^2\theta=1-\cos^2\theta=1-\frac{1}{9}=\frac{8}{9}$
$\displaystyle \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=\frac{\frac{8}{9}}{\frac{1}{9}}=8$
$\displaystyle \therefore \frac{6\sin^2\theta+\tan^2\theta}{4\cos\theta}$
$\displaystyle =\frac{6\left(\frac{8}{9}\right)+8}{4\left(\frac{1}{3}\right)}$
$\displaystyle =\frac{\frac{16}{3}+8}{\frac{4}{3}}$
$\displaystyle =\frac{\frac{40}{3}}{\frac{4}{3}}=10$
$\displaystyle \therefore \text{The required value is }10.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }\sqrt{3}\tan\theta=3\sin\theta,\text{ find the value of }\sin^2\theta-\cos^2\theta. \hfill\text{[CBSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{3}\tan\theta=3\sin\theta$
$\displaystyle \sqrt{3}\frac{\sin\theta}{\cos\theta}=3\sin\theta$
$\displaystyle \therefore \frac{\sqrt{3}}{\cos\theta}=3$
$\displaystyle \therefore \cos\theta=\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \cos^2\theta=\frac{1}{3}$
$\displaystyle \sin^2\theta=1-\cos^2\theta=1-\frac{1}{3}=\frac{2}{3}$
$\displaystyle \therefore \sin^2\theta-\cos^2\theta=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$
$\displaystyle \therefore \text{The required value is }\frac{1}{3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }\mathrm{cosec}\theta=\frac{13}{12},\text{ find the value of } \frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}.\hfill\text{[CBSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \mathrm{cosec}\theta=\frac{13}{12}$
$\displaystyle \therefore \sin\theta=\frac{12}{13}$
$\displaystyle \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{144}{169}}=\frac{5}{13}$
$\displaystyle \therefore \frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}$
$\displaystyle =\frac{2\left(\frac{12}{13}\right)-3\left(\frac{5}{13}\right)}{4\left(\frac{12}{13}\right)-9\left(\frac{5}{13}\right)}$
$\displaystyle =\frac{\frac{24-15}{13}}{\frac{48-45}{13}}$
$\displaystyle =\frac{9}{3}=3$
$\displaystyle \therefore \text{The required value is }3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If }\sin\theta+\cos\theta=\sqrt{2}\sin\theta,\text{ find }\cot\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin\theta+\cos\theta=\sqrt{2}\sin\theta$
$\displaystyle \cos\theta=(\sqrt{2}-1)\sin\theta$
$\displaystyle \therefore \frac{\cos\theta}{\sin\theta}=\sqrt{2}-1$
$\displaystyle \therefore \cot\theta=\sqrt{2}-1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{If }2\sin^2\theta-\cos^2\theta=2,\text{ find the value of }\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle 2\sin^2\theta-\cos^2\theta=2$
$\displaystyle 2\sin^2\theta-(1-\sin^2\theta)=2$
$\displaystyle \qquad[\because \cos^2\theta=1-\sin^2\theta]$
$\displaystyle 3\sin^2\theta-1=2$
$\displaystyle 3\sin^2\theta=3$
$\displaystyle \sin^2\theta=1$
$\displaystyle \therefore \sin\theta=1$
$\displaystyle \therefore \theta=90^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{If }\sqrt{3}\tan\theta-1=0,\text{ find the value of }\sin^2\theta-\cos^2\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{3}\tan\theta-1=0$
$\displaystyle \therefore \tan\theta=\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \tan^2\theta=\frac{1}{3}$
$\displaystyle \sec^2\theta=1+\tan^2\theta=1+\frac{1}{3}=\frac{4}{3}$
$\displaystyle \therefore \cos^2\theta=\frac{1}{\sec^2\theta}=\frac{3}{4}$
$\displaystyle \sin^2\theta=1-\cos^2\theta=1-\frac{3}{4}=\frac{1}{4}$
$\displaystyle \therefore \sin^2\theta-\cos^2\theta=\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}$
$\displaystyle \therefore \text{The required value is }-\frac{1}{2}.$
$\displaystyle \\$