Prove the identity $\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B}=\sin B+\cos B$ I have worked on this identity from both sides of the equation and can't seem to get it to equal the other side no matter what I try.
$\displaystyle\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B} =\sin B+\cos B$

HINT:
$\displaystyle\frac{\cos B}{1-\tan B}=\frac{\cos B}{1-\dfrac{\sin B}{\cos B}}=\frac{\cos^2B}{\cos B-\sin B}$
$\displaystyle\frac{\sin B}{1-\cot B}=\frac{\sin B}{1-\dfrac{\cos B}{\sin B}}=\frac{\sin^2B}{\sin B-\cos B}=-\frac{\sin^2B}{\cos B-\sin B}$
