Question

If ${\sin }^{-1}a+{\sin }^{-1}b+{\sin }^{-1}c=\pi$, then $a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}=mabc$, then the value of m =

Answer 1

${\sin }^{-1}a+{\sin }^{-1}b+{\sin }^{-1}c=\pi$

$a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}=m\left(abc\right)$

${\sin }^{-1}a=x,s{\in }^{-1}b=y,{\sin }^{-1}c=z$

$a=\sin x,b=\sin y,c=\sin z$

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

$=\sin x\cos x+\sin y\cos y+\sin z\cos z$

$=\frac{1}{2}(\sin 2x+\sin 2y+\sin 2z$

As $x+y+z=\pi$,

$\sin 2x+\sin 2y+\sin 2z=4\sin x\sin y\sin z$

$=2\sin x\sin y\sin 2$

$=2abc$

$\therefore m=2$

$\therefore$ The correct answer is 2.