Question

If ${\sin }^{-1}x+{\sin }^{-1}(1-x)={\sin }^{-1}\sqrt{1-x^2}$, then $x=$

• A. $0,\frac{1}{2}$
• B. $0,-\frac{1}{2}$
• C. $-\frac{1}{2},\frac{1}{2}$
• D. none of these

Answer 1

The correct option is A

$0,\frac{1}{2}$

Explanation for the correct option.

Step 1: Simplify the expersion

Given that, ${\sin }^{-1}x+{\sin }^{-1}(1-x)={\sin }^{-1}\sqrt{1-x^2}$

Put $x=\sin \theta$

$\begin{array}{rcl}{\sin }^{-1}\left(\sin \theta \right)+{\sin }^{-1}(1-\sin \theta )& =& {\sin }^{-1}\sqrt{1-{\sin }^2\theta }\\ \theta +{\sin }^{-1}(1-\sin \theta )& =& {\sin }^{-1}\left(\cos \theta \right)\\ \theta +{\sin }^{-1}(1-\sin \theta )& =& {\sin }^{-1}\sin \left(\frac{\mathrm{\pi }}{2}-\theta \right)\left[\cos \theta =\sin \left(\frac{\mathrm{\pi }}{2}-\theta \right)\right]\\ \theta +{\sin }^{-1}(1-\sin \theta )& =& \left(\frac{\mathrm{\pi }}{2}-\theta \right)\\ {\sin }^{-1}(1-\sin \theta )& =& \frac{\mathrm{\pi }}{2}-2\theta \end{array}$

Step 2: Find $x$

Taking $\sin$ both sides we have:

$\begin{array}{rcl}1-\sin \theta & =& \sin \left(\frac{\mathrm{\pi }}{2}-2\theta \right)\\ 1-\sin \theta & =& \cos 2\theta \\ 1-\sin \theta & =& 1-2{\sin }^2\theta \\ 2{\sin }^2\theta -\sin \theta & =& 0\\ \sin \theta \left(2\sin \theta -1\right)& =& 0\end{array}$

$\begin{array}{rcl}& \Rightarrow & 2\sin \theta -1=0or\sin \theta =0\\ & \Rightarrow & \sin \theta =\frac{1}{2},0\\ & \Rightarrow & x=0,\frac{1}{2}\end{array}$

Hence, option A is correct.