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.
Hint:
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…
