Question

If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\pi ,$ then prove that:
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

Answer 1

Let ${\sin }^{-1}x=A\Rightarrow \sin A=x$

${\sin }^{-1}y=B\Rightarrow \sin B=y$

${\sin }^{-1}z=C\Rightarrow \sin C=z$

Given, ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\pi$

$\Rightarrow A+B+C=p\Rightarrow 2A+2B+2C=2\pi$

So, using trigonometric property,

$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$

We can write,

$\Rightarrow 2\sin A\cos A+2\sin B\cos B+2\sin C\cos C=4\sin A\sin B\sin C$

$\Rightarrow 2\sin A.\sqrt{1-{\sin }^{2}A}+2\sin B.\sqrt{1-{\sin }^{2}B}+\sin C.\sqrt{1-{\sin }^{2}C}=4\sin A\sin B\sin C$

$2x\sqrt{1-x^2}+2y\sqrt{1-y^2}+2z\sqrt{1-z^2}=4xyz$

$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

Hence proved