Question

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

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

Answer 1

The correct option is A $2abc$

Let ${\sin }^{-1}a=A,{\sin }^{-1}b=B,{\sin }^{-1}c=C$
$\Rightarrow \sin A=a,\sin B=b,\sin C=c$
Given $A+B+C=\pi$, which is possible iff $A,B,C\in [0,\frac{\pi }{2}]$

Now,
$a\sqrt{(1-a^2)}+b\sqrt{{(1-b)}^2}+c\sqrt{(1-c^2)}$
$=\sin A\sqrt{(1-{\sin }^{2}A)}+\sin B\sqrt{(1-{\sin }^{2}B)}+\sin C\sqrt{(1-{\sin }^{2}C)}$
$=\sin A\cos A+\sin B\cos B+\sin C\cos C(\because A,B,C\in [0,\frac{\pi }{2}])$
$=\frac{1}{2}(\sin 2A+\sin 2B+\sin 2C)$
$=2\sin A\sin B\sin C=2abc$