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
$si{n}^{-1}\left(sinA\right)+si{n}^{-1}\left(sinB\right)+si{n}^{-1}\left(sinC\right)=\pi$
Where
$a=sinA$, $b=sinB$ and $c=sinC$
Hence $A,B$ and $C$ can be assumed as angles of triangle.
Therefore.
$a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}$
Simplifies as
$sinAcosA+sinBcosB+sinCcosC$.

Multiplying and dividing by $2$, we get

$\frac{sin2A+sin2B+sin2C}{2}$
$=\frac{2sin(A+B)cos(A-B)+sin2C}{2}$
$=\frac{2sinCcos(A-B)+2sinCcosC}{2}$ ...$(A+B=\pi -C)$
$=sinC\left(cos\right(A-B)+cos(\pi -(A+B)\left)\right)$
$=sinC\left(cos\right(A-B)-cos(A+B\left)\right)$
$=2sinCsinAsinB$
$=2abc$