#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 5 Maths Textbook Solution.

$3$
Given:
$\left ( \sin^{-1}x \right )^{2}+\left (\sin^{-1}y \right )^{2}+\left ( \sin^{-1}z \right )^{2}= \frac{3}{4}\pi ^{2} \,\, then\, find \,\, x^{2}+y^{2}+z^{2}.$
Hint:
$We\, know\, that \, the\, range\, o\! f \, \sin^{-1} is \left [ \frac{-\pi }{2} ,\frac{\pi }{2}\right ]$
$\sin^{-1}x\leq \frac{\pi }{2},\sin^{-1}y \leq \frac{\pi}{2}, \sin^{-1}z\leq \frac{\pi}{2}$
$\left ( \sin^{-1}x \right )^{2}\leq \frac{\pi ^{2}}{4},\left (\sin^{-1}y \right )^{2}\leq \frac{\pi ^{2}}{4},\left ( \sin^{-1}z \right )^{2}\leq \frac{\pi ^{2}}{4}$
$\left ( \sin^{-1}x \right )^{2}+\left (\sin^{-1}y \right )^{2}+\left ( \sin^{-1}z \right )^{2}\leq \frac{3}{4}\pi ^{2} \, \, and \, it\, is\, given\, that\, L.H.S\, is \,\, \frac{3}{4}\pi ^{2}.$
$So\: it\: is\: possible\: only\: when$
$\sin^{-1}x= \frac{\pi }{2},\sin^{-1}y= \frac{\pi}{2}, \sin^{-1}z= \frac{\pi}{2}$
$x= 1,y= 1,z= 1$
$\therefore x^{2}+y^{2}+z^{2}= 3$