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

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)$ = $tan^{-1}\: tan\large \frac{x}{2}=\frac{x}{2}$

edited Mar 14, 2013