# Thread: Verify the following identity

1. ## Verify the following identity

Okay so I'm trying to verify the following identity: $\frac{\tan^2\theta-1}{3\tan^2\theta+2\tan\theta-1}=\frac{\tan\theta-1}{3\tan\theta-1}$

Working on the right hand side I did this: $\frac{\tan\theta-1}{3\tan\theta-1}*\frac{\tan\theta+1}{\tan\theta+1}$

and that works out to be: $\frac{\tan^2\theta-1}{3\tan^2\theta+2\tan\theta-1}$

Am I even allowed to do what I did?

2. Dear iamanoobatmath,

By the second line you have written you could get the right hand side by cancellation.

$\frac{\tan\theta-1}{3\tan\theta-1}*\frac{\tan\theta+1}{\tan\theta+1}=\frac{\tan\th eta-1}{3\tan\theta-1}$

3. I have no idea how to solve the identity and when I try to work on the left side this ends up happening. Could you show me how to verify the identity?

4. Originally Posted by iamanoobatmath
I have no idea how to solve the identity and when I try to work on the left side this ends up happening. Could you show me how to verify the identity?
Sudaharaka told you how to do it.

Factorise the top and the bottom and cancel the common factor.

$\frac{\tan^2{x} - 1}{3\tan^2{x} + 2\tan{x} - 1}$

$= \frac{(\tan{x} + 1)(\tan{x} - 1)}{(3\tan{x} - 1)(\tan{x} + 1)}$

$= \frac{\tan{x} - 1}{3\tan{x} - 1}$.