# Thread: Please check proof of identity

1. ## Please check proof of identity

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.

2. ## Re: Please check proof of identity

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.
That way is fine. I'd have omitted the third line of working since the reader should be able to establish that you converted the existing $\displaystyle 1-\sin^2 \theta$ into $\displaystyle \cos^2 \theta$

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

3. ## Re: Please check proof of identity

Originally Posted by e^(i*pi)
That way is fine. I'd have omitted the third line of working since the reader should be able to establish that you converted the existing $\displaystyle 1-\sin^2 \theta$ into $\displaystyle \cos^2 \theta$

Spoiler:
This one -- $\displaystyle \equiv\dfrac{2\sin\theta\cos\theta}{1 + \cos^2\theta - (1 - cos^2\theta)}$
Thank you very much indeed, I didn't see that. I keep missing simple things like that. Thank you for checking my work and pointing out that step.