Question

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

Answer 1

We are given,

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

let,

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

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

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

here,

$A,B,C\in [-\pi /2,\pi /2]$

$\therefore A+B+C=\pi$ ........ $\left(1\right)$

now,

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

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

$=\sin A.\cos A+\sin B\cos B+\sin C.\cos C$

[Multiply & Divide by $2$]

$=\frac{2}{2}[\sin A\cos A+\sin B\cos B]+\sin C.\cos C$

$=\frac{[2\sin A\cos A+2\sin B\cos B]}{2}+\sin C.\cos C(2\sin \theta \cos \theta =\sin 2\theta )$

$=\frac{(\sin A+\sin 2B)}{2}+\sin C.\cos C$

$=\frac{2\sin \left(\frac{2A+2B}{2}\right)\cos \left(\frac{2A-2B}{2}\right)}{2}+\sin C.\cos C$

$=\sin (A+B)\cos (A-B)+\sin C.\cos C$

$=\sin (\pi -C)\cos (A-B)+\sin C.\cos (\pi -(A+B\left)\right)$

$=\sin C\left[\cos \right(A-B)-\cos (A+B\left)\right]$

$=\sin C[-2\sin \left(\frac{-2B}{2}\right)\sin \left(\frac{3A}{2}\right)]$

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

$=2xyz$

$=RHS$.