Question

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

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

Answer 1

Answer: (A)

$2abc$

The above expression can be re-written as
${\sin }^{-1}\left(\sin A\right)+{\sin }^{-1}\left(\sin B\right)+{\sin }^{-1}\left(\sin C\right)=\pi$

Where
$a=\sin A$, $b=\sin B$ and $c=\sin C$

Hence $A,B$ and $C$ can be assumed as angles of the triangle.

Therefore.
$a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}$

Simplifies as
$\sin A\cos A+\sin B\cos B+\sin C\cos C$.

Multiplying and dividing by 2, we get
$\Rightarrow \frac{\sin 2A+\sin 2B+\sin 2C}{2}$

$=\frac{2\sin (A+B)\cos (A-B)+\sin 2C}{2}$

$=\frac{2\sin C\cos (A-B)+2\sin C\cos C}{2}$ ...$(A+B=\pi -C)$

$=\sin C\left(\cos \right(A-B)+\cos (\pi -(A+B)\left)\right)$

$=\sin C\left(\cos \right(A-B)-\cos (A+B\left)\right)$

$=2\sin C\sin A\sin B$

$=2abc$