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

To Prove: $\sin^{-1}\frac{63}{65}= \sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
Hint:  Using R.H.S term, here first we will we will convert $\sin^{-1}x$ and $\cos^{-1}x$ then use formula.
$\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}\right ]$
Solution: Taking R.H.S
R.H.S = $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
=$= \sin^{-1}\frac{5}{13}+\sin^{-1}\frac{4}{5} \; \; \; \; \; \; \; \; \; [\because \cos^{-1}x=\sin^{-1}\sqrt{1-x^{2}}]$
Then we will use the formula
$\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}\right ]$
\begin{aligned} &= \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) \end{aligned}
$= \sin^{-1}\left ( \frac{63}{65} \right )$
$= L.H.S$
Hence $\sin^{-1}\frac{63}{65}= \sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$