1. ## Identity problem

Prove that the left equals the right
$1-(sin^2x)/(1+cotx)-(cos^2x)/(1+tanx) = sin2x/2$

So far I've multiplied to get a common denominator of $1+tanx+cotx+cotxtanx$ And have the numerator of $cotx+tanx+cotxtanx$ Past this I can't think of where to go.

2. I will use s & c for sine and cosine respectively.

First convert the LHS to sines and cosines:

$1 - \frac{s^2}{1+c/s} - \frac{c^2}{1 + s/c}$

Now simplify a bit:

$1 - \frac{s^3}{s+c} - \frac{c^3}{s+c}$

Put everything into one fraction and use the sum of cubes factorization:

$\frac{(s+c) - (s+c)(s^2-sc+c^2)}{s+c}$

The (s+c) terms cancel leaving

$1 - (1 - sc) = sc = \frac{2sc}{2}$

And now you are one identity from completing the problem.

3. thanks for the help, took me a while to understand the steps you took but I got it now. You even explained it in less steps then my teacher.