Question

$y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$

Prove that: $y=2{\cos }^{-1}x$

Answer 1

Consider the given equation,

$y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$

Put, $x=\cos \theta \Rightarrow \theta ={\cos }^{-1}x$

$y={\sec }^{-1}\left(\frac{1}{2{\cos }^2\theta -1}\right)$

$y={\sec }^{-1}\left(\frac{1}{\cos 2\theta }\right)$

$y={\sec }^{-1}\sec 2\theta$

$y=2\theta$

$y=2{\cos }^{-1}x$

Hence, this is the answer.