Question

The value of $si{n}^{-1}[cos\{co{s}^{-1}(cosx)+si{n}^{-1}(sinx)\}]$ where $x\in (\frac{\pi }{2},\pi )$ is equal to

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

Answer 1

Answer: (D)

$si{n}^{-1}[cos\{co{s}^{-1}(cosx)+si{n}^{-1}(sinx)\}]$
$=si{n}^{-1}[cos(x+\pi -x)]$ as $x\in (\pi /2,\pi )$
$=si{n}^{-1}(cos\pi )=si{n}^{-1}(-1)=-\frac{\pi }{2}$