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

Answer 1

The correct option is A

$2abc$

Explanation for the correct option.

Step 1: Relation between angle

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

Put $a=\sin A,b=\sin B\text{and}c=\sin C$

$$\begin{gathered} \Rightarrow {\sin }^{-1}\left(\sin A\right)+{\sin }^{-1}\left(\sin B\right)+{\sin }^{-1}\left(\sin C\right)=\pi \\ \Rightarrow A+B+C=\mathrm{\pi } \end{gathered}$$

Step 2: Find the value of the expression

So,$a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}$ can be rewritten as:
$a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}=\sin A\cos A+\sin B\cos B+\sin C\cos C$
Multiply and divide by 2, on RHS we get
$\begin{array}{rcl}a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}& =& \frac{\sin 2\mathrm{A}+\sin 2\mathrm{B}+\sin 2\mathrm{C}}{2}\\ & =& \frac{2\sin (A+B)\cos (A-B)+\sin 2C}{2}\\ & =& \frac{2\sin C\cos (A-B)+2\sin C\cos C}{2}\dots (A+B=\pi -C)\\ & =& \sin C\left(\cos \right(A-B)+\cos (\pi -(A+B)\left)\right)\\ & =& \sin C\left(\cos \right(A-B)-\cos (A+B\left)\right)\\ & =& 2\sin C\sin A\sin B\\ & =& 2abc\end{array}$

Hence, option A is correct.