Question

If $xϵ[-1,0)$, then $\pi -\left(co{s}^{-1}\right(2x^2-1)-2si{n}^{-1}x)$ is equal to

• A. $-\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{3\pi }{2}$
• D. $-2\pi$

Answer 1

The correct option is B $\pi$

$xϵ[-1,0)$
$\pi -{\cos }^{-1}(2x^2-1)-(-2{\sin }^{-1}(x\left)\right)$
$=\pi -{\cos }^{-1}(2x^2-1)+2{\sin }^{-1}\left(x\right)$
Let $x=\sin A$
$=\pi -{\cos }^{-1}\left(\cos 2A\right)+2{\sin }^{-1}\left(sinA\right)$
$=\pi -2A+2A$
$=\pi$