Question

The value of ${\sin }^{-1}\left[\cos \left\{{\cos }^{-1}\right(\mathrm{cosx})+{\sin }^{-1}(\sin x\left)\right\}\right]$ where $x\in \left(\frac{\pi }{2},\pi \right)$ is equal to

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

Answer 1

The correct option is D $-\frac{\pi }{2}$

Solution -

$si{n}^{-1}\left[cos\right[co{s}^{-1}\left(cosx\right)+si{n}^{-1}\left(sinx\right)\left]\right]$

$x\in (\frac{\pi }{2},\pi )$

we know that Range of $si{n}^{-1}(-)$ is from $[\frac{-\pi }{2},\frac{\pi }{2}]$

Also range of $co{s}^{-1}(-)$ is from $[0,\pi ]$

For $xϵ(\frac{\pi }{2},\pi )\to si{n}^{-1}sinx=\pi -x$

$co{s}^{-1}cosx=x$

$=si{n}^{-1}\left[cos\right(\pi -x+x\left)\right]$

$=si{n}^{-1}\left[cos\pi \right]$

$=si{n}^{-1}[-1]$

$=\frac{-\pi }{2}$

D is correct.