#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 3 Maths Textbook Solution.

Answer:$x= \frac{\sqrt{3-1}}{2\sqrt{2}}\: and\: y= \frac{1}{\sqrt{2}}$
Hint: Changing $\sin^{-1}x$ to$\cos^{-1}x$ and vice versa; We know that,
$\sin^{-1}x+\cos^{-1}x= \frac{\pi }{2}$
Given:$\cos^{-1}x-\cos^{-1}y$  $= \frac{\pi }{6}$  and$\sin^{-1}x+ \sin^{-1}y = \frac{\pi }{3}$
Here, we have to compute x  and y.
Solution:
Let’s replace $\cos^{-1}x-\cos^{-1}y$  by $\sin^{-1}x\: and\: \sin^{-1}y$ , therefore
$\! \! \! \! \! \! \! \! \Rightarrow \cos^{-1}x-\cos^{-1}y = \frac{\pi }{6} \cdot \cdot \cdot (1)\\ \Rightarrow \sin^{-1}x+ \sin^{-1}y = \frac{\pi }{3} \cdot \cdot \cdot \cdot (2)\\ \Rightarrow \frac{\pi }{2} - \sin^{-1}x - \frac{\pi }{2} + \sin^{-1}y = \frac{\pi }{6}\\ -(\sin^{-1}x-\sin^{-1}y) = \frac{\pi }{6} \cdot \cdot \cdot (3)$
Subtracting Equation 3 from 2,
$\! \! \! \! \! \! \! \! \! \sin^{-1}x+\sin^{-1}y +\sin^{-1}x-\sin^{-1}y = \frac{\pi }{3} - \frac{\pi }{6}\\ 2 \sin^{-1}x = \frac{\pi }{6}\\ \sin^{-1}x = \frac{\pi }{12}\\ x = \sin \frac{\pi }{12}\\$
$x= \frac{\sqrt{3}-1}{2\sqrt{2}}$
$\! \! \! \! \! \! \! \! \! 2 \sin^{-1}y = \frac{\pi }{3} + \frac{\pi }{6}\\ 2 \sin^{-1}y = \frac{\pi }{2}\\ \sin^{-1}y = \frac{\pi }{4}\\ y = \sin \frac{\pi }{4}\\ y = \frac{1}{\sqrt2}$
Concept: Properties and relations between inverse trigonometric functions.
Note: Values of Trigonometric functions for various radians and degrees.