# Thread: Another Compound Trig Identity

1. ## Another Compound Trig Identity

I'm not sure if what I did was right, so please verify it.

sin 2x/(1-cos 2x)=2 csc 2x-tan x

This is what I did:

LS= sin 2x/(1-cos2x)
= (2sinx cosx)/(1-(cos^2 x-sin^2 x))
= (2sinx cosx)/(cos^2 x+sin^2 x-cos^2 x+sin^2 x)
= (2 sin x cos x)/(2sin^2 x)
= cot x

RS= 2/(2sinx cosx)-sinx/cosx
= (2-2sin^2 x)/(2sinx cosx)
= (1-sin^2 x)/(sinx cosx)
= (1-(1-cos^2 x))/(sinx cosx)
= cos^2 x/(sinx cosx)
= cot x

2. Hello, DemonX01!

However, you should not work with both sides of the identity.

I did find a way to change one side into the other . . .

Verify: . $\frac{\sin 2x}{1-\cos2x} \:=\:2\csc 2x-\tan x$
I started with the right side . . .

$2\csc2x - \tan x \;\;=\;\;\frac{2}{\sin2x} - \frac{\sin x}{\cos x} \;\;=\;\;\frac{2}{2\sin x\cos x} \;-\; \frac{\sin x}{\cos x}$

. . . . . $=\;\;\frac{1}{\sin x\cos x} - \frac{\sin x}{\cos x} \;\;=\;\;\frac{1}{\sin x\cos x} - \frac{\sin x}{\cos x}\cdot{\color{blue}\frac{\sin x}{\sin x}}$

. . . . . $=\;\;\frac{1-\sin^2\!x}{\sin x\cos x} \;\;=\;\;\frac{\cos^2\!x}{\sin x\cos x} \;\;=\;\;\frac{\cos x}{\sin x }$

Multiply by $\frac{2\sin x}{2\sin x}\!:\;\;\frac{2\sin x}{2\sin x}\cdot\frac{\cos x}{\sin x} \;=\;\frac{2\sin x\cos x}{2\sin^2\!x} \;=\;\frac{\sin2x}{1-\cos 2x}\;\;\begin{array}{c} \\ {\color{red}**} \end{array}$

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**

The denomniator comes from the identity:

. . $\sin^2\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 2\sin^2\!\theta \:=\:1 - \cos2\theta$