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

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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 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 

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