Trigonometry functions How do I verify the following identity:
$$\frac{1-(\sin x - \cos x)^2}{\sin x} = 2\cos x$$
I have done simpler problems but got stuck with this one.
Please help.
Tony

$$\frac{1-(\sin x -\cos x)^2}{\sin x}=\frac{1-(\sin^2x+\cos^2x)+2\sin x\cos x}{\sin x}=\frac{2\cos x\sin x}{\sin x}=2\cos x$$
Here I used the identity: $\sin^2x+\cos^2x=1$
