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Question:

If $\left(\sin ^{-1} x\right)^{2}+\left(\sin ^{-1} y\right)^{2}+\left(\sin ^{-1} z\right)^{2}=\frac{3}{4} \pi^{2}$, find the value of $x^{2}+y^{2}+z^{2}$

Solution:

We know that the maximum value of $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ is $\frac{\pi}{2}$ and minimum value of $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ is $-\frac{\pi}{2}$

Now,

For maximum value

$\mathrm{LHS}=\left(\sin ^{-1} x\right)^{2}+\left(\sin ^{-1} y\right)^{2}+\left(\sin ^{-1} z\right)^{2}$

$=\left(\frac{\pi}{2}\right)^{2}+\left(\frac{\pi}{2}\right)^{2}+\left(\frac{\pi}{2}\right)^{2}$

$=\frac{3}{4} \pi^{2}=\mathrm{RHS}$

and For minimum value

$\mathrm{LHS}=\left(\sin ^{-1} x\right)^{2}+\left(\sin ^{-1} y\right)^{2}+\left(\sin ^{-1} z\right)^{2}$

$=\left(-\frac{\pi}{2}\right)^{2}+\left(-\frac{\pi}{2}\right)^{2}+\left(-\frac{\pi}{2}\right)^{2}$

$=\frac{3}{4} \pi^{2}=$ RHS

Now, For maximum value

$\sin ^{-1} x=\frac{\pi}{2}, \sin ^{-1} y=\frac{\pi}{2}, \sin ^{-1} z=\frac{\pi}{2}$

$\Rightarrow x=\sin \frac{\pi}{2}, y=\sin \frac{\pi}{2}, z=\sin \frac{\pi}{2}$

$\Rightarrow x=1, y=1, z=1$

$\therefore x^{2}+y^{2}+z^{2}=1+1+1=3$

and for minimum value

$\sin ^{-1} x=-\frac{\pi}{2}, \sin ^{-1} y=-\frac{\pi}{2}, \sin ^{-1} z=-\frac{\pi}{2}$

$\Rightarrow x=\sin \left(-\frac{\pi}{2}\right), y=\sin \left(-\frac{\pi}{2}\right), z=\sin \left(-\frac{\pi}{2}\right)$

$\Rightarrow x=-1, y=-1, z=-1$

$\therefore x^{2}+y^{2}+z^{2}=1+1+1=3$