Class 12: Inverse Trigonometric Functions

$\displaystyle \textbf{Question 1: }\text{Evaluate each of the following:}$
$\displaystyle \text{(i) }\sin\left(\sin^{-1}\frac{7}{25}\right)\qquad \text{(ii) }\sin\left(\cos^{-1}\frac{5}{13}\right)$
$\displaystyle \text{(iii) }\sin\left(\tan^{-1}\frac{24}{7}\right)\qquad \text{(iv) }\sin\left(\sec^{-1}\frac{17}{8}\right)$
$\displaystyle \text{(v) }\mathrm{cosec}\left(\cos^{-1}\frac{3}{5}\right)\qquad \text{(vi) }\sec\left(\sin^{-1}\frac{12}{13}\right)$
$\displaystyle \text{(vii) }\tan\left(\cos^{-1}\frac{8}{17}\right)\qquad \text{(viii) }\cot\left(\cos^{-1}\frac{3}{5}\right)$
$\displaystyle \text{(ix) }\cos\left(\tan^{-1}\frac{24}{7}\right)$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{(ii) }\sin\left(\cos^{-1}\frac{5}{13}\right)$
$\displaystyle =\sin\left(\sin^{-1}\sqrt{1-\left(\frac{5}{13}\right)^2}\right)$
$\displaystyle =\sin\left(\sin^{-1}\sqrt{1-\frac{25}{169}}\right)$
$\displaystyle =\sin\left(\sin^{-1}\sqrt{\frac{144}{169}}\right)$
$\displaystyle =\sin\left(\sin^{-1}\frac{12}{13}\right)=\frac{12}{13}$

$\displaystyle \text{(iii) }\sin\left(\tan^{-1}\frac{24}{7}\right)$
$\displaystyle =\sin\left(\sin^{-1}\frac{\frac{24}{7}}{\sqrt{1+\left(\frac{24}{7}\right)^2}}\right)$
$\displaystyle =\sin\left(\sin^{-1}\frac{\frac{24}{7}}{\sqrt{1+\frac{576}{49}}}\right)$
$\displaystyle =\sin\left(\sin^{-1}\frac{\frac{24}{7}}{\sqrt{\frac{625}{49}}}\right)$
$\displaystyle =\sin\left(\sin^{-1}\frac{\frac{24}{7}}{\frac{25}{7}}\right)=\frac{24}{25}$

$\displaystyle \text{(iv) }\sin\left(\sec^{-1}\frac{17}{8}\right)$
$\displaystyle =\sin\left(\cos^{-1}\frac{8}{17}\right)$
$\displaystyle =\sin\left(\sin^{-1}\sqrt{1-\left(\frac{8}{17}\right)^2}\right)$
$\displaystyle =\sin\left(\sin^{-1}\sqrt{1-\frac{64}{289}}\right)$
$\displaystyle =\sin\left(\sin^{-1}\sqrt{\frac{225}{289}}\right)$
$\displaystyle =\sin\left(\sin^{-1}\frac{15}{17}\right)=\frac{15}{17}$

$\displaystyle \text{(v) }\mathrm{cosec}\left(\cos^{-1}\frac{3}{5}\right)$
$\displaystyle =\mathrm{cosec}\left(\sin^{-1}\sqrt{1-\left(\frac{3}{5}\right)^2}\right)$
$\displaystyle =\mathrm{cosec}\left(\sin^{-1}\sqrt{1-\frac{9}{25}}\right)$
$\displaystyle =\mathrm{cosec}\left(\sin^{-1}\sqrt{\frac{16}{25}}\right)$
$\displaystyle =\mathrm{cosec}\left(\sin^{-1}\frac{4}{5}\right)$
$\displaystyle =\mathrm{cosec}\left(\mathrm{cosec}^{-1}\frac{5}{4}\right)=\frac{5}{4}$

$\displaystyle \text{(vi) }\sec\left(\sin^{-1}\frac{12}{13}\right)$
$\displaystyle =\sec\left(\cos^{-1}\sqrt{1-\left(\frac{12}{13}\right)^2}\right)$
$\displaystyle =\sec\left(\cos^{-1}\sqrt{1-\frac{144}{169}}\right)$
$\displaystyle =\sec\left(\cos^{-1}\sqrt{\frac{25}{169}}\right)$
$\displaystyle =\sec\left(\cos^{-1}\frac{5}{13}\right)$
$\displaystyle =\sec\left(\sec^{-1}\frac{13}{5}\right)=\frac{13}{5}$

$\displaystyle \text{(vii) }\tan\left(\cos^{-1}\frac{8}{17}\right)$
$\displaystyle =\tan\left(\tan^{-1}\frac{\sqrt{1-\left(\frac{8}{17}\right)^2}}{\frac{8}{17}}\right)$
$\displaystyle =\tan\left(\tan^{-1}\frac{\frac{15}{17}}{\frac{8}{17}}\right)=\frac{15}{8}$

$\displaystyle \text{(viii) }\cot\left(\cos^{-1}\frac{3}{5}\right)$
$\displaystyle =\cot\left(\tan^{-1}\frac{\sqrt{1-\left(\frac{3}{5}\right)^2}}{\frac{3}{5}}\right)$
$\displaystyle =\cot\left(\tan^{-1}\frac{\frac{4}{5}}{\frac{3}{5}}\right)$
$\displaystyle =\cot\left(\cot^{-1}\frac{3}{4}\right)=\frac{3}{4}$
$\displaystyle \\$

$\displaystyle \text{(ix) }\cos\left(\tan^{-1}\frac{24}{7}\right)$
$\displaystyle =\cos\left(\cos^{-1}\frac{1}{\sqrt{1+\left(\frac{24}{7}\right)^2}}\right)$
$\displaystyle =\cos\left(\cos^{-1}\frac{1}{\sqrt{1+\frac{576}{49}}}\right)$
$\displaystyle =\cos\left(\cos^{-1}\frac{1}{\sqrt{\frac{625}{49}}}\right)$
$\displaystyle =\cos\left(\cos^{-1}\frac{1}{\frac{25}{7}}\right)=\frac{7}{25}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Prove the following results:}$
$\displaystyle \text{(i) }\tan\left(\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{2}{3}\right)=\frac{17}{6}$
$\displaystyle \text{(ii) }\cos\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\right)=\frac{6}{5\sqrt{13}}\qquad \textbf{[CBSE 2012]}$
$\displaystyle \text{(iii) }\tan\left(\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}\right)=\frac{63}{16}$
$\displaystyle \text{(iv) }\sin\left(\cos^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\right)=\frac{63}{65}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) LHS}=\tan\left(\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{2}{3}\right)$
$\displaystyle =\frac{\tan\left(\cos^{-1}\frac{4}{5}\right)+\tan\left(\tan^{-1}\frac{2}{3}\right)}{1-\tan\left(\cos^{-1}\frac{4}{5}\right)\tan\left(\tan^{-1}\frac{2}{3}\right)}$
$\displaystyle =\frac{\tan\left(\tan^{-1}\frac{3}{4}\right)+\tan\left(\tan^{-1}\frac{2}{3}\right)}{1-\tan\left(\tan^{-1}\frac{3}{4}\right)\tan\left(\tan^{-1}\frac{2}{3}\right)}$
$\displaystyle =\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times\frac{2}{3}}$
$\displaystyle =\frac{\frac{17}{12}}{\frac{1}{2}}=\frac{17}{6}=\text{RHS. Hence, proved.}$

$\displaystyle \text{(ii) LHS}=\cos\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\right)$
$\displaystyle =\cos\left(\sin^{-1}\frac{3}{5}\right)\cos\left(\cot^{-1}\frac{3}{2}\right)-\sin\left(\sin^{-1}\frac{3}{5}\right)\sin\left(\cot^{-1}\frac{3}{2}\right)$
$\displaystyle =\cos\left(\cos^{-1}\frac{4}{5}\right)\cos\left(\cos^{-1}\frac{3}{\sqrt{13}}\right)-\sin\left(\sin^{-1}\frac{3}{5}\right)\sin\left(\sin^{-1}\frac{2}{\sqrt{13}}\right)$
$\displaystyle =\frac{4}{5}\times\frac{3}{\sqrt{13}}-\frac{3}{5}\times\frac{2}{\sqrt{13}}$
$\displaystyle =\frac{12-6}{5\sqrt{13}}=\frac{6}{5\sqrt{13}}=\text{RHS. Hence, proved.}$

$\displaystyle \text{(iii) LHS}=\tan\left(\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}\right)$
$\displaystyle =\frac{\tan\left(\sin^{-1}\frac{5}{13}\right)+\tan\left(\cos^{-1}\frac{3}{5}\right)}{1-\tan\left(\sin^{-1}\frac{5}{13}\right)\tan\left(\cos^{-1}\frac{3}{5}\right)}$
$\displaystyle =\frac{\tan\left(\tan^{-1}\frac{5}{12}\right)+\tan\left(\tan^{-1}\frac{4}{3}\right)}{1-\tan\left(\tan^{-1}\frac{5}{12}\right)\tan\left(\tan^{-1}\frac{4}{3}\right)}$
$\displaystyle =\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times\frac{4}{3}}$
$\displaystyle =\frac{\frac{7}{4}}{\frac{4}{9}}=\frac{63}{16}=\text{RHS. Hence, proved.}$

$\displaystyle \text{(iv) LHS}=\sin\left(\cos^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\right)$
$\displaystyle =\sin\left(\cos^{-1}\frac{3}{5}\right)\cos\left(\sin^{-1}\frac{5}{13}\right)+\cos\left(\cos^{-1}\frac{3}{5}\right)\sin\left(\sin^{-1}\frac{5}{13}\right)$
$\displaystyle =\sin\left(\sin^{-1}\frac{4}{5}\right)\cos\left(\cos^{-1}\frac{12}{13}\right)+\cos\left(\cos^{-1}\frac{3}{5}\right)\sin\left(\sin^{-1}\frac{5}{13}\right)$
$\displaystyle =\frac{4}{5}\times\frac{12}{13}+\frac{3}{5}\times\frac{5}{13}$
$\displaystyle =\frac{48+15}{65}=\frac{63}{65}=\text{RHS. Hence, proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve: }\cos\left(\sin^{-1}x\right)=\frac{1}{6}$
$\displaystyle \text{Answer:}$
$\displaystyle \cos\left(\sin^{-1}x\right)=\frac{1}{6}$
$\displaystyle \Rightarrow \cos\left(\cos^{-1}\sqrt{1-x^2}\right)=\frac{1}{6}$
$\displaystyle \Rightarrow \sqrt{1-x^2}=\frac{1}{6}$
$\displaystyle \Rightarrow 1-x^2=\frac{1}{36}$
$\displaystyle \Rightarrow x^2=1-\frac{1}{36}=\frac{35}{36}$
$\displaystyle \Rightarrow x=\pm\frac{\sqrt{35}}{6}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Solve: }\cos\left(2\sin^{-1}(-x)\right)=0$
$\displaystyle \text{Answer:}$
$\displaystyle \cos\left(2\sin^{-1}(-x)\right)=0$
$\displaystyle \Rightarrow \cos^2\left(\sin^{-1}(-x)\right)-\sin^2\left(\sin^{-1}(-x)\right)=0$
$\displaystyle \Rightarrow \cos^2\left(\cos^{-1}\sqrt{1-x^2}\right)-\sin^2\left(\sin^{-1}(-x)\right)=0$
$\displaystyle \Rightarrow \left(\sqrt{1-x^2}\right)^2-(-x)^2=0$
$\displaystyle \Rightarrow 1-x^2-x^2=0$
$\displaystyle \Rightarrow 1-2x^2=0$
$\displaystyle \Rightarrow x^2=\frac{1}{2}$
$\displaystyle \Rightarrow x=\pm\frac{1}{\sqrt2}$
$\displaystyle \\$