# Prove that sec 2x + tan 2x = tan (x + pi/4)

I've been trying to figure out how to do this question, and no matter what I do, I can't seem to the solution. Below is my working so far,

LHS =

= sec 2x + tan 2x = 1/cos2x + sin2x/cos2x = 1+sin2x/cos2x

I've tried double angle (albeit, most likely incorrectly, if I keep failing to reach a solution). I've also tried comp. angles etc.

Any help would be greatly appreciated!

The left-hand side can be written as $$\frac{1+\sin(2x)}{\cos(2x)}$$ the right-hand side as $$\frac{\sin(x)+\cos(x)}{\cos(x)-\sin(x)}$$ multiply numerator and denominator of the last term by $$\sin(x)+\cos(x)$$
We expand the RHS to get $$\frac{\tan x + \tan \frac{\pi}{4}}{1 - \tan x \tan \frac{\pi}{4}} = \frac{\tan x + 1}{1 - \tan x} = \frac{\frac{\sin x}{\cos x}+1}{1 - \frac{\sin x}{\cos x}} = \frac{\cos x + \sin x}{\cos x - \sin x},$$ and we now multiply top and bottom by $$\cos x + \sin x$$ to obtain $$\frac{\cos^2 x + \sin^2 x + 2 \sin x \cos x}{\cos^2 x - \sin^2 x} = \frac{1+2 \sin x \cos x}{\cos^2 x - \sin^2 x}=\frac{1+\sin 2x}{\cos 2x},$$ as required.
Set $$t=\tan x$$ and use the double-angle formulæ (the first one is valid for $$x\not\equiv \pm\frac\pi 4\bmod\pi$$): $$\tan 2x=\frac{2t}{1-t^2},\qquad \cos 2x=\frac{1-t^2}{1+t^2}.$$
Hint: set $$x+\pi/4=y$$, so the left-hand side becomes $$\frac{1+\sin(2y-\pi/2)}{\cos(2y-\pi/2)}=\frac{1-\cos2y}{\sin2y}$$ Now use the duplication formulas…