Hi all,

I am having trouble with the following question. I have filled a few pages with various manipulations and I am no closer to proving the identity. I have obviously missed a trick. Any help would be appreciated.

First the question:

Prove that $\displaystyle \frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} \equiv \sin \theta + \cos \theta$

Let's take the right hand side and see what we can do with it:

$\displaystyle \frac{\sin \theta}{1} + \frac{\cos \theta}{1}$

$\displaystyle \frac{\sin^2 \theta}{\sin \theta} + \frac{\cos^2 \theta}{\cos \theta}$

$\displaystyle \frac{1 - \cos^2 \theta}{\sin \theta} + \frac{1 - \sin^2 \theta}{\cos \theta}$

$\displaystyle \frac{1}{\sin \theta} - \frac{\cos^2 \theta}{\sin \theta} + \frac{1}{\cos \theta} - \frac{\sin^2 \theta}{\cos \theta}$

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

$\displaystyle \frac{\sin \theta -\sin \theta \cos^2 \theta}{1-\cos^2 \theta} + \frac{\cos \theta - \cos \theta \sin^2 \theta}{1 - \sin^2 \theta}$

I can't really see where to go with it.

Regards,

Evanator