Question

If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{\pi }{2},$ then the value of $x^2+y^2+z^2+2xyz$ is equal to

Answer 1

We have,
${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{\pi }{2}$
Let ${\sin }^{-1}x=\alpha ,{\sin }^{-1}y=\beta$ and ${\sin }^{-1}z=\gamma \cdots \left(1\right)$
$\therefore \alpha +\beta +\gamma =\frac{\pi }{2}$
$\alpha +\beta =\frac{\pi }{2}-\gamma$
$\cos (\alpha +\beta )=\cos (\frac{\pi }{2}-\gamma )$
$\Rightarrow \cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta =\sin \gamma \cdots \left(2\right)$
From $\left(1\right),$
$\sin \alpha =x$
$\Rightarrow \cos \alpha =\sqrt{1-x^2}$
Similarly, $\cos \beta =\sqrt{1-y^2}$

Now, from equation $\left(2\right),$
$\sqrt{1-x^2}\sqrt{1-y^2}=xy+z$
Squaring both sides, we have
$(1-x^2)(1-y^2)=x^2{y}^2+z^2+2xyz$
$\Rightarrow 1-y^2-x^2+x^2{y}^2=x^2{y}^2+z^2+2xyz$
$\Rightarrow x^2+y^2+z^2+2xyz=1$