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Write the following function in the simplest form: \[ tan^{-1} \bigg( \sqrt {\frac{1-cos \: x}{1 + cos \: x}} \bigg), x < \pi\]

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Toolbox:
  • \( 1-cos\theta = 2sin^2\frac{\theta}{2}\)
  • \( 1+cos\theta = 2cos^2\frac{\theta}{2}\)
  • \( tan \theta = \large \frac{ \sin \theta}{\cos \theta} \)
Given $\; tan^{-1} \bigg( \sqrt {\large \frac{1-cos \: x}{1 + cos \: x}}$$ \bigg), x < \pi$:
Substituting for \( 1-cos\theta = 2sin^2\frac{\theta}{2}\) and \( 1+cos\theta = 2cos^2\frac{\theta}{2}\), we can further reduce this:
\(\Rightarrow tan^{-1} \bigg( \sqrt{\large\frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}} \bigg) \)\(=tan^{-1}\bigg(\large \frac{sin\frac{x}{2}}{cos\frac{x}{2}}\bigg)\)
Substituting \( tan \theta = \large \frac{ \sin \theta}{\cos \theta} \), this reduces to:
$\; tan^{-1} \bigg( \sqrt {\large \frac{1-cos \: x}{1 + cos \: x}}$$ \bigg)$ = \( tan^{-1}\: tan\large \frac{x}{2}=\frac{x}{2}\)

 

answered Feb 22, 2013 by thanvigandhi_1
edited Mar 14, 2013 by balaji.thirumalai
 

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