# 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