Question:
If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z+\sin ^{-1} t=2 \pi$, then find the value of $x^{2}+y^{2}+z^{2}+t^{2}$
Solution:
We know that the maximum value of $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ and $\sin ^{-1} t$ is $\frac{\pi}{2}$
Now,
$\mathrm{LHS}=\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z+\sin ^{-1} t$
$=\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}$
$=2 \pi=\mathrm{RHS}$
Now,
$\sin ^{-1} x=\frac{\pi}{2}, \sin ^{-1} y=\frac{\pi}{2}, \sin ^{-1} z=\frac{\pi}{2}$ and $\sin ^{-1} t=\frac{\pi}{2}$
$\Rightarrow x=\sin \frac{\pi}{2}, y=\sin \frac{\pi}{2}, z=\sin \frac{\pi}{2}$ and $t=\sin \frac{\pi}{2}$
$\Rightarrow x=1, y=1, z=1$ and $t=1$
$\therefore x^{2}+y^{2}+z^{2}+t^{2}=1+1+1+1=4$