# How do you prove tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)?

Oct 5, 2016

see below

#### Explanation:

${\tan}^{2} \left(\frac{1}{2} \theta\right) = \frac{\tan \theta - \sin \theta}{\tan \theta + \sin \theta}$

Right Side $= \frac{\tan \theta - \sin \theta}{\tan \theta + \sin \theta}$

$= \frac{\sin \frac{\theta}{\cos} \theta - \sin \theta}{\sin \frac{\theta}{\cos} \theta + \sin \theta}$

$= \frac{\frac{\sin \theta - \sin \theta \cos \theta}{\cos} \theta}{\frac{\sin \theta + \sin \theta \cos \theta}{\cos} \theta}$

$= \frac{\sin \theta - \sin \theta \cos \theta}{\cos} \theta \cdot \cos \frac{\theta}{\sin \theta + \sin \theta \cos \theta}$

$= \frac{\sin \theta - \sin \theta \cos \theta}{\sin \theta + \sin \theta \cos \theta}$

$= \frac{\sin \theta \left(1 - \cos \theta\right)}{\sin \theta \left(1 + \cos \theta\right)}$

$= \frac{1 - \cos \theta}{1 + \cos \theta}$

$= {\left(\tan \left(\frac{1}{2} \theta\right)\right)}^{2}$

$= {\tan}^{2} \left(\frac{1}{2} \theta\right)$

$=$ Left Side