Question

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

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

Answer 1

The correct option is A

$2\mathrm{abc}$

Explanation for the correct option.

Step 1: Given information

Given that, ${\sin }^{-1}a+{\sin }^{-1}b+{\sin }^{-1}c=\pi$.

Put $a=\sin A,b=\sin Bandc=\sin C$

$\Rightarrow A+B+C=\mathrm{\pi }$
Hence $A,B$ and $C$ can be assumed to be angles of the triangle.
Therefore, $a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}$

on substituting values gives:

$$\begin{gathered} =\sin A\sqrt{1-{\sin }^{2}A}+\sin B\sqrt{1-{\sin }^{2}B}+\sin C\sqrt{1-{\sin }^{2}C} \\ =\sin A\cos A+\sin B\cos B+\sin C\cos C\left[\because \sqrt{1-{\sin }^2\theta }=\cos \theta \right] \end{gathered}$$

Step 2: Multiplying and dividing by 2, we get

$\sin A\cos A+\sin B\cos B+\sin C\cos C=\frac{\sin 2\mathrm{A}+\sin 2\mathrm{B}+\sin 2\mathrm{C}}{2}[\because \sin 2x=2\sin x\cos x]$
$\begin{array}{rcl}\therefore \frac{\sin 2\mathrm{A}+\sin 2\mathrm{B}+\sin 2\mathrm{C}}{2}& =& \frac{2\sin (A+B)\cos (A-B)+\sin 2C}{2}\left[[\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)\right]\\ & =& \frac{2\sin C\cos (A-B)+2\sin C\cos C}{2}(\because A+B=\pi -C)\\ & =& \sin C\left(\cos \right(A-B)+\cos (\pi -(A+B)\left)\right)\\ & =& \sin C\left(\cos \right(A-B)-c\mathrm{os}(A+B\left)\right)\\ & =& 2\sin C\sin A\sin B\\ & =& 2abc\end{array}$

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

Hence, option A is correct.