Browse Questions

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

Toolbox:
• $sec^2\theta-1=tan^2\theta$
• $cot\theta = tan ( \large\frac{\pi}{2}-\theta)$
• $\large \large\frac{1}{tan \theta}$ = $cot \theta$
Given, $tan^{-1} \large \frac{1}{\sqrt{x^2 - 1}}$, $| x | > 1$:

We need to remove the square root sign to help simplify the identity:

Let $\;x = sec\theta \Rightarrow \theta = sec^{-1}x$. Substiting for $x$ and using known trignometric identities, we can further reduce this:

Substituting, $tan^{-1}\big(\large \frac{1}{\sqrt{x^2-1}}\big)\:$ becomes$\: tan^{-1} \bigg( \large \frac{1}{\sqrt{sec^2\theta-1}} \bigg)$
Substituting for $sec^2\theta-1=tan^2\theta$, this reduces to $tan^{-1}\large \frac{1}{tan\theta}$

Substituting for $\large \frac{1}{tan \theta}$ = $cot \theta$, this reduces to $tan^{-1}(cot\theta)$
Substituting for $cot\theta = tan (\large \frac{\pi}{2}-\theta)$, we get $tan^{-1}\: tan \bigg( \large\frac{\pi}{2}-\theta \bigg)$

$tan^{-1} \large \frac{1}{\sqrt{x^2 - 1}}$ $= \large\frac{\pi}{2}-\theta=\large\frac{\pi}{2}-sec^{-1}x$

edited Mar 15, 2013