Question

If ${\sec }^{-1}x=cose{c}^{-1}y$, then find the value of ${\cos }^{-1}\frac{1}{x}+{\cos }^{-1}\frac{1}{y}$

Answer 1

Given equation ${\sec }^{-1}x={\csc }^{-1}y$

Now Taking

$={\cos }^{-1}\frac{1}{x}+{\cos }^{-1}\frac{1}{y}$ [using property ${\cos }^{-1}\left(\frac{1}{x}\right)={\sec }^{-1}x$]

$={\sec }^{-1}x+{\sec }^{-1}y$ Putting ${\sec }^{-1}x={\csc }^{-1}y$

$={\csc }^{-1}y+{\sec }^{-1}y$

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