-1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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 $$\sin(x)+\cos(x)$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Hint:

Set $t=\tan x$ and use the double-angle formulæ (the first one is valid for $x\not\equiv \pm\frac\pi 4\bmod\pi$): $$\tan 2x=\frac{2t}{1-t^2},\qquad \cos 2x=\frac{1-t^2}{1+t^2}.$$

$\endgroup$
0
$\begingroup$

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…

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.