Originally Posted by

**Furyan** Hello all

The question is show that:

$\displaystyle \dfrac{\sin 2\theta}{1+\cos 2\theta} \equiv \tan\theta$

I have

$\displaystyle LHS \equiv\dfrac{\sin\theta\cos\theta + \cos\theta\sin\theta}{1 + \cos^2\theta - \sin^2\theta}$

$\displaystyle \equiv\dfrac{2\sin\theta\cos\theta}{1 + \cos^2\theta - (1 - cos^2\theta)}$

$\displaystyle \equiv\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}$

$\displaystyle \equiv\dfrac{\sin\theta}{\cos\theta} \equiv \tan\theta $

Is that alright and is the a better way?

Thank you.