Question

If ${\sin }^{-1}\left(a\right)+{\sin }^{-1}\left(b\right)+{\sin }^{-1}\left(c\right)=\pi$ 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.

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

Let, $a=\sin \left(x\right),b=\sin \left(y\right),c=sin\left(z\right)$

$\Rightarrow x+y+z=\mathrm{\pi }....\left(1\right)$

Step 2: Evaluate $a\sqrt{\left(1-a^2\right)}+b\sqrt{\left(1-b^2\right)}+c\sqrt{\left(1-c^2\right)}$

Substitute values of $a,b,c$ in the expression.

$\begin{array}{rcl}\therefore a\sqrt{\left(1-a^2\right)}+b\sqrt{\left(1-b^2\right)}+c\sqrt{\left(1-c^2\right)}& =& \sin x\sqrt{1-{\sin }^{2}x}+\sin y\sqrt{1-{\sin }^{2}y}+\sin z\sqrt{1-{\sin }^{2}z}\\ & =& \sin \left(x\right)\cos \left(x\right)+\sin \left(y\right)\cos \left(y\right)+\sin \left(z\right)\cos \left(z\right)\\ & =& \frac{1}{2}\left[2\sin \left(x\right)\cos \left(x\right)+2\sin \left(y\right)\cos \left(y\right)+2\sin \left(z\right)\cos \left(z\right)\right]\\ & =& \frac{1}{2}\left[\sin \left(2x\right)+\sin \left(2y\right)+\sin \left(2z\right)\right]...\left(2\right)\end{array}$

Step 3: Simplify equation (2)

Now using identity, $\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$ we have:

$\begin{array}{rcl}& \Rightarrow & \frac{1}{2}\left[\sin 2x+\sin 2y+\sin 2z\right]=\frac{1}{2}\left[2\sin \left(\frac{2x+2y}{2}\right)\cos \left(\frac{2x-2y}{2}\right)+\sin 2z\right]\\ & =& \frac{1}{2}\left[2\sin \left(x+y\right)\cos \left(x-y\right)+\sin 2z\right]\\ & =& \frac{1}{2}\left[2\sin \left(\mathrm{\pi }-\mathrm{z}\right)\cos \right(x-y)+2\sin z\cos z]\\ & =& \sin z\left[\cos \left(x-y\right)+\cos z\right]\\ & =& \sin z\left[\cos \left(x-y\right)+\cos \left\{\mathrm{\pi }-\left(\mathrm{x}+\mathrm{y}\right)\right\}\right]\\ & =& \sin z\left[\cos \left(x-y\right)+\cos \left(\mathrm{x}+\mathrm{y}\right)\right]\\ & =& \sin z\left[2\sin \left(x\right)\sin \left(y\right)\right]\\ & =& 2abc\end{array}$

Hence, option A is correct.