Question

Write the following in the simplest form.

$ta{n}^{-1}\frac{1}{\sqrt{x^2-1}},|x|>1$

Answer 1

Let $x=sec\theta ,then\theta =se{c}^{-1}x$

$So,ta{n}^{-1}\frac{1}{\sqrt{x^2-1}}=ta{n}^{-1}\left(\frac{1}{\sqrt{se{c}^2\theta -1}}\right)=ta{n}^{-1}\left(\frac{1}{\sqrt{ta{n}^2\theta }}\right)$

$(\because se{c}^2\theta -ta{n}^2\theta =1)$

$=ta{n}^{-1}\left(\frac{1}{tan\theta }\right)=ta{n}^{-1}\left(cot\theta \right)=ta{n}^{-1}\left(tan(\frac{\pi }{2}-\theta )\right)(\because tan(\frac{\pi }{2}-\theta )=cot\theta )$

$=\frac{\pi }{2}-\theta =\frac{\pi }{2}-se{c}^{-1}x$