1. ## Identity Proof!

tan( pi/4 + x/2 ) = sec x + tan x

I have worked on this for over two hours now.
Help would be greatly appreciated

Thanks!

2. Originally Posted by dxscorp
tan( pi/4 + x/2 ) = sec x + tan x

I have worked on this for over two hours now.
Help would be greatly appreciated

Thanks!
using this
$\tan (a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

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

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

can you continue?, you must know

$\cos 2x = \cos ^2 x - \sin ^2 x \Longrightarrow \cos x = \cos ^2 \left(\frac{x}{2}\right) - \sin ^2 \left(\frac{x}{2}\right)$

$\sin 2x = 2\sin x \cos x \Longrightarrow \sin x = 2 \sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)$

3. Hello, dxscorp!

$\tan\left(\frac{\pi}{4}+ \frac{x}{2}\right) \:=\: \sec x + \tan x$

$\tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \;=\;\frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}} \;=\;\frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}$ . $=\;\frac{1 + \dfrac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} {1 - \dfrac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} \;=\;\frac{\cos\frac{x}{2} + \sin\frac{x}{2}} {\cos\frac{x}{2} - \sin\frac{x}{2}}$

Multiply by $\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\!:$

. . $\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} \cdot {\color{blue}\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}}$

. . . . . . $=\;\;\frac{\overbrace{(\cos^2\tfrac{x}{2} + \sin^2\tfrac{x}{2})}^{\text{This is 1}} \;+\; \overbrace{2\sin\tfrac{x}{2}\cos\tfrac{x}{2}}^{\te xt{This is }\sin x}} {\underbrace{\cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2}}_{\text{This is }\cos x}}$

. . . . . . $=\;\;\frac{1 + \sin x}{\cos x} \;\;=\;\;\frac{1}{\cos x} + \frac{\sin x}{\cos x} \;\;=\;\; \sec x + \tan x$