Trignometric identities - Prove that the following is an identity.

Quote:

Originally Posted by Kakariki

Hello, I am having troubles with a question involving trigonometric identities. I have always had trouble with these types of problems, so I need as much help as possible.
To the problem:

"Prove that the following is an identity: $\frac{cos2\theta}{1 + sin2\theta} = \frac{cot \theta - 1}{cot \theta +1}$ "

Here is my attempt:

R.H.S
$\frac{\frac{1}{tan\theta} - 1}{\frac{1}{tan\theta} + 1}$
$\frac{\frac{cos\theta}{sin\theta} -1}{\frac{cos\theta}{sin\theta} +1}$

I don't know what to do from here. On the left hand start I can't see where to even start. Any help at all is very much appreciated. I really do not understand these problems.

Thank you.

$\frac{\cot{\theta} - 1}{\cot{\theta} + 1} = \frac{\frac{\cos{\theta}}{\sin{\theta}} - 1}{\frac{\cos{\theta}}{\sin{\theta}} + 1}$

$= \frac{\frac{\cos{\theta} - \sin{\theta}}{\sin{\theta}}}{\frac{\cos{\theta} + \sin{\theta}}{\sin{\theta}}}$

$= \frac{\cos{\theta} - \sin{\theta}}{\cos{\theta} + \sin{\theta}}$

$= \frac{(\cos{\theta} - \sin{\theta})^2}{(\cos{\theta} + \sin{\theta})(\cos{\theta} - \sin{\theta})}$

$= \frac{\cos^2{\theta} - 2\cos{\theta}\sin{\theta} + \sin^2{\theta}}{\cos^2{\theta} - \sin^2{\theta}}$

$= \frac{1 - \sin{(2\theta)}}{\cos{(2\theta)}}$

$= \frac{[1 - \sin{(2\theta)}][1 + \sin{(2\theta)}]}{\cos{(2\theta)}[1 + \sin{(2\theta)}]}$

$= \frac{1 - \sin^2{(2\theta)}}{\cos{(2\theta)}[1 + \sin{(2\theta)}]}$

$= \frac{\cos^2{(2\theta)}}{\cos{(2\theta)}[1 + \sin{(2\theta)}]}$

$= \frac{\cos{(2\theta)}}{1 + \sin{(2\theta)}}$ .