Question

Write ${\cot }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right),x>1$ in the simplest form.

Answer 1

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