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

## 1 Answer

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 $$tan^{-1}\: tan \bigg( \large\frac{\pi}{2}-\theta \bigg)$$

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

answered Feb 22, 2013
edited Mar 15, 2013

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