Question

If ${\sin }^{-1}a+{\sin }^{-1}b+{\sin }^{-1}c=\pi$, then the value of $a\cdot \sqrt{(1-a^2)}+b\cdot \sqrt{(1-b^2)}+c\cdot \sqrt{(1-c^2)}$ will be

• A. $2abc$
• B. $abc$
• C. $\frac{1}{2}abc$
• D. $\frac{1}{3}abc$

Answer 1

Answer: (A)

$2abc$

Let ${\sin }^{-1}a=A$,
${\sin }^{-1}b=B$
and ${\sin }^{-1}C=C$
Therefore, $\sin A=a,\sin B=b,\sin C=C$
and $A+B+C=\pi$
Then $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$
$\Rightarrow \sin A\cos A+\sin B\cos B+\sin C\cos C=2\sin A\sin B\sin C$
$\Rightarrow \sin A\sqrt{1-{\sin }^{2}A}+\sin B\sqrt{1-{\sin }^{2}B}+\sin C\sqrt{1-{\sin }^{2}C}=2\sin A\sin B\sin C$
$\Rightarrow a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}=2abc$,
while ${\sin }^{-1}a+{\sin }^{-1}b+{\sin }^{-1}c=\pi$.