Question

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

Answer 1

Given: ${\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right),|x|>1$
Substituting $x=\sec \theta$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{{\sec }^2\theta -1}}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{(1+{\tan }^2\theta )-1}}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{{\tan }^2\theta }}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)={\tan }^{-1}\left(\frac{1}{\tan \theta }\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)={\tan }^{-1}\left(\cot \theta \right)$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)={\tan }^{-1}\left(\tan (\frac{\pi }{2}-\theta )\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\frac{\pi }{2}-\theta$
$\therefore x=\sec \theta$
$\Rightarrow \theta ={\sec }^{-1}x$
So,
$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\frac{\pi }{2}-\theta =\frac{\pi }{2}-{\sec }^{-1}x$