Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

First, simplify each term:

${\sin }^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}={\sin }^{-1}\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)=\frac{\pi }{12}$

${\cos }^{-1}\frac{\sqrt{12}}{4}={\cos }^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{6}$

${\sec }^{-1}\sqrt{2}=\frac{\pi }{4}$.

Then, $\frac{\pi }{12}+\frac{\pi }{6}+\frac{\pi }{4}=\frac{\pi +2\pi +3\pi }{12}=\frac{6\pi }{12}=\frac{\pi }{2}$.

So, ${\sin }^{-1}[\cot (\frac{\pi }{2})]={\sin }^{-1}[0]=0$.