Question

Write the function in the simplest form:
${\tan }^{-1}\frac{1}{\sqrt{x^2-1}},\left|x\right|>1$

Answer 1

${\tan }^{-1}\frac{1}{\sqrt{x^2-1}},\left|x\right|>1$

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

$\therefore {\tan }^{-1}\frac{1}{\sqrt{x^2-1}}={\tan }^{-1}\frac{1}{\sqrt{{cosec}^2\theta -1}}$

$={\tan }^{-1}\left(\frac{1}{\cot \theta }\right)={\tan }^{-1}\left(\tan \theta \right)$

$=\theta ={cosec}^{-1}x$

$=\frac{\pi }{2}-{\sec }^{-1}x$ ...$[\because {cosec}^{-1}x+{\sec }^{-1}x=\frac{\pi }{2}]$