# Question f8601

May 3, 2016

RHS=LHS proves the identity

#### Explanation:

To prove the identity, it may be simpler to start with RHS.

RHS= $\frac{\tan \left(\frac{\pi}{4}\right) + \tan \left(\frac{A}{2}\right)}{1 - \tan \left(\frac{\pi}{4}\right) \tan \left(\frac{A}{2}\right)}$ =(1+tan(A/2))/(1-tan(A/2)#

= $\frac{\cos \left(\frac{A}{2}\right) + \sin \left(\frac{A}{2}\right)}{\cos \left(\frac{A}{2}\right) - \sin \left(\frac{A}{2}\right)}$ =$\frac{\cos \left(\frac{A}{2}\right) + \sin \left(\frac{A}{2}\right)}{\cos \left(\frac{A}{2}\right) - \sin \left(\frac{A}{2}\right)} \cdot \frac{\cos \left(\frac{A}{2}\right) + \sin \left(\frac{A}{2}\right)}{\cos \left(\frac{A}{2}\right) + \sin \left(\frac{A}{2}\right)}$

=$\frac{{\cos}^{2} \left(\frac{A}{2}\right) + {\sin}^{2} \left(\frac{A}{2}\right) + 2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)}{{\cos}^{2} \left(\frac{A}{2}\right) - {\sin}^{2} \left(\frac{A}{2}\right)}$

$= \frac{1 + \sin A}{\cos} A = \sec A + \tan A = L H S$