Question

The value of $\cos ({\sin }^{-1}\frac{1}{2}+{\sec }^{-1}2)$.

Answer 1

Now,

$\cos ({\sin }^{-1}\frac{1}{2}+{\sec }^{-1}2)$

$=\cos ({\sin }^{-1}\frac{1}{2}+{\cos }^{-1}\frac{1}{2})$

$=\cos \frac{\pi }{2}$ [ Since ${\sin }^{-1}\frac{1}{2}+{\cos }^{-1}\frac{1}{2}=\frac{\pi }{2}$]

$=0$.