# Prove that tan(pi/4+A/2)=secA+tanA?

May 3, 2016

#### Explanation:

Let us start from Right Hand Side

$\tan \left(\frac{\pi}{4} + \frac{A}{2}\right)$

= (tan(pi/4)+tan(A/2))/(1-tan((pi)/4)tan(A/2)

= (1+tan(A/2))/(1-1*tan(A/2)

= (1+sin(A/2)/cos(A/2))/(1-sin(A/2)/cos(A/2)

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

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

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

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

= $\frac{1 + \sin A}{\cos} A$

= $\frac{1}{\cos} A + \sin \frac{A}{\cos} A$

= $\sec A + \tan A$