Question

The solution set of the equation ${\cos }^{-1}x-{\sin }^{-1}x={\sin }^{-1}(1-x)$ is

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

Answer 1

The correct option is B $[0,\frac{1}{2}]$

${\cos }^{-1}x-{\sin }^{-1}x={\sin }^{-1}(1-x)$

$si{n}^{-1}x+co{s}^{-1}x=\pi /2$
$\frac{\pi }{2}-2{\sin }^{-1}x={\sin }^{-1}(1-x)$
$\Rightarrow 2{\sin }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}(1-x)$
$\Rightarrow 2{\sin }^{-1}x={\cos }^{-1}(1-x)$
$\Rightarrow 2{\cos }^{-1}\sqrt{1-x^2}={\cos }^{-1}(1-x)$
Put $x=\sin \theta$
$\Rightarrow 2\theta ={\cos }^{-1}(1-\sin \theta )$
$\Rightarrow \cos 2\theta =1-\sin \theta$
$\Rightarrow 2{\sin }^2\theta -\sin \theta =0$
$\Rightarrow \sin \theta =0or\sin \theta =\frac{1}{2}$
$\Rightarrow x=0or\frac{1}{2}$