#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 2 Subquestion (ii) Maths Textbook Solution.

To prove:  $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}= \tan^{-1}\frac{63}{16}$
Hint: First we will convert $\sin^{-1}x$ and $\cos^{-1}x$ then we will use the formula of
$\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}} \right ]$
Solution: Taking L.H.S
\begin{aligned} \text { L.H.S } &=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5} \\ &=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \sqrt{1-\left(\frac{3}{5}\right)^{2}} \quad\left[\because \cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}}\right] \\ &=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{4}{5} \end{aligned}
Now, using the formula we have,
\begin{aligned} \text { L. } H . S &=\sin ^{-1}\left[\frac{5}{13} \sqrt{1-\left(\frac{4}{5}\right)^{2}}+\frac{4}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}\right] \\ &=\sin ^{-1}\left(\frac{5}{13} \times \frac{3}{5}+\frac{4}{5} \times \frac{12}{13}\right) \\ &=\sin ^{-1}\left(\frac{15}{65}+\frac{48}{65}\right) \\ &=\sin ^{-1}\left(\frac{63}{65}\right) \end{aligned}
\begin{aligned} &=\tan ^{-1}\left(\frac{63 / 65}{1-\left(\frac{63}{65}\right)^{2}}\right) \quad\left[\sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}}\right] \\ \end{aligned}
$= \tan^{-1}\left ( \frac{\frac{63}{65}}{\frac{16}{65}} \right )$
$\tan^{-1}\left ( \frac{63}{16} \right )= R.H.S$
Hence  $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}= \tan^{-1}\frac{63}{16}$