Question

If ${\tan }^{-1}(x^2+3|x|-4)+{\cot }^{-1}(4\pi +{\sin }^{-1}(\sin 14\left)\right)=\frac{\pi }{2}$, then the value of ${\sin }^{-1}\left(\sin 2|x|\right)$ is equal to.

• A. $6-2\pi$
• B. $2\pi -6$
• C. $\pi -3$
• D. $3-\pi$

Answer 1

Answer: (A)

$6-2\pi$

${\sin }^{-1}\left(\sin 14\right)=(14-4\pi )$

Since,

${\tan }^{-1}(x^2+3|x|-4)+{\cot }^{-1}(4\pi +{\sin }^{-1}(\sin 14\left)\right)=\frac{\pi }{2}$

or, ${\tan }^{-1}(x^2+3|x|-4)={\tan }^{-1}(4\pi +{\sin }^{-1}(\sin 14\left)\right)$
or, $x^2+3|x|-4=4\pi +(14-4\pi )$
$\Rightarrow x^2+3|x|-18=0$
$\Rightarrow (|x|+6)(|x|-3)=0$
$\therefore |x|=3$
So, ${\sin }^{-1}\left(\sin 2|x|\right)={\sin }^{-1}\left(\sin 6\right)=(6-2\pi )$.