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

To prove: $\frac{9\pi}{8}-\frac{9}{4}\sin ^{-1}\frac{1}{3}= \frac{9}{4}\sin^{-1}\left ( \frac{2\sqrt{2}}{3} \right )$
Hint: First we take the common in the question then we proceed.
Solution: Taking L.H.S.
L.H.S = $\frac{9 \pi}{8}-\frac{9}{4}\sin ^{-1}\frac{1}{3}$
= $\frac{9}4\left [ \frac{\pi }{2}-\sin^{-1}\frac{1}{3}\right ]$
=  $\frac{9}{4}\left [ \cos^{-1}\left ( \frac{1}{3} \right ) \right ]\; \; \; \; \left [ \because \frac{\pi }{2}-\sin^{-1}x=\cos^{-1}x \right ]$
=  $\frac{9}{4}\left [ \sin^{-1}\left ( 1-\sqrt{\left ( \frac{1}{3} \right )^{2}} \right ) \right ]\; \; \; \; \; \; \left [ \because \cos^{-1}x= \sin^{-1}\sqrt{1-x^{2}} \right ]$
$= \frac{9}{4}\sin^{-1}\sqrt{\frac{9-1}{9}}$
$= \frac{9}{4}\sin^{-1}\left ( \frac{2\sqrt{2}}{3} \right )$
= R.H.S
Hence,
$\frac{9\pi}{8}-\frac{9}{4}\sin ^{-1}\frac{1}{3}= \frac{9}{4}\sin^{-1}\left ( \frac{2\sqrt{2}}{3} \right )$