If $\;$ $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi$, $\;$ then prove that $\;$ $x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2} = 2 x y z$
$\left\{ \text{Note: } \sin A + \sin B = 2 \sin \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A - B}{2}\right) \right.$
$\left. \cos A - \cos B = 2 \sin \left(\dfrac{A + B}{2}\right) \sin \left(\dfrac{B - A}{2}\right) \right\}$
Let $\;$ $\sin^{-1} x = \alpha$, $\;$ $\sin^{-1} y = \beta$, $\;$ $\sin^{-1} z = \gamma$, $\;\;$ $\alpha, \; \beta, \; \gamma \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
Then, $\;$ $\sin \alpha = x$, $\;$ $\sin \beta = y$, $\;$ $\sin \gamma = z$ $\;\;\; \cdots \; (1)$
and $\;$ $\cos \alpha = \sqrt{1 - x^2}$, $\;$ $\cos \beta = \sqrt{1 - y^2}$, $\;$ $\cos \gamma = \sqrt{1 - z^2}$ $\;\;\; \cdots \; (2)$
Given: $\;$ $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi$
$\implies$ $\alpha + \beta + \gamma = \pi$ $\;\;\; \cdots \; (3)$
Now,
$\begin{aligned}
\sin \left(2 \alpha\right) + \sin \left(2 \beta\right) + \sin \left(2 \gamma\right) & = 2 \sin \left(\dfrac{2 \alpha + 2 \beta}{2}\right) \cos \left(\dfrac{2 \alpha - 2 \beta}{2}\right) + 2 \sin \gamma \cos \gamma \\\\
& = 2 \sin \left(\alpha + \beta\right) \cos \left(\alpha - \beta\right) + 2 \sin \gamma \cos \gamma \\\\
& = 2 \sin \left(\pi - \gamma\right) \cos \left(\alpha - \beta\right) + 2 \sin \gamma \cos \gamma \\
& \hspace{2cm} \left[\text{In view of equation } (3)\right] \\\\
& = 2 \sin \gamma \cos \left(\alpha - \beta\right) + 2 \sin \gamma \cos \gamma \\\\
& = 2 \sin \gamma \left[\cos \left(\alpha - \beta\right) + \cos \gamma\right] \\\\
& = 2 \sin \gamma \left\{ \cos \left(\alpha - \beta\right) + \cos \left[\pi - \left(\alpha + \beta\right)\right] \right\} \\
& \hspace{2cm} \left[\text{In view of equation } (3)\right] \\\\
& = 2 \sin \gamma \left[\cos \left(\alpha - \beta\right) - \cos \left(\alpha + \beta\right)\right] \\\\
& = 2 \sin \gamma \times 2 \sin \left(\dfrac{\alpha - \beta + \alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha + \beta - \alpha + \beta}{2}\right) \\\\
& = 4 \sin \alpha \times \sin \beta \times \sin \gamma
\end{aligned}$
i.e. $\;$ $2 \sin \alpha \cos \alpha + 2 \sin \beta \cos \beta + 2 \sin \gamma \cos \gamma = 4 \sin \alpha \times \sin \beta \times \sin \gamma$
i.e. $\;$ $\sin \alpha \cos \alpha + \sin \beta \cos \beta + \sin \gamma \cos \gamma = 2 \sin \alpha \times \sin \beta \times \sin \gamma$ $\;\;\; \cdots \; (4)$
$\therefore \;$ In view of equations $(1)$ and $(2)$, equation $(4)$ becomes,
$x \sqrt{1 - x^2} + y \sqrt{1 - y^2} + z \sqrt{1 - z^2} = 2 x y z$
Hence proved.