Question

Simplify ${\cot }^{-1}\frac{1}{\sqrt{x^2-1}}$ for $x<-1$

• A. ${\cos }^{-1}x$
• B. ${\sec }^{-1}x$
• C. ${\text{cosec}}^{-1}x$
• D. ${\tan }^{-1}x$

Answer 1

The correct option is B ${\sec }^{-1}x$

${\cot }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)$

$letx={\sec }^2\theta$

${\cot }^{-1}\left(\frac{1}{\sqrt{{\sec }^2\theta -1}}\right)$

$1+{\tan }^2\theta ={\sec }^2\theta$

$\therefore {\sec }^2\theta -1={\tan }^2\theta$

${\cot }^{-1}\left(\frac{1}{\sqrt{{\tan }^2\theta }}\right)$

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

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

$={\sec }^{-1}x.$