Originally Posted by
Chop Suey Is this your question? $\displaystyle \frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}} + 1 = 2\sin{x}$
I'm going to assume that the left hand side is written correctly, but I'm going to ask you to recheck the right hand side.
After this step:
$\displaystyle \frac{\frac{\sin{x}}{\cos{x}} - \frac{\cos{x}}{\sin{x}}}{\frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}} \cdot \frac{\sin{x}\cos{x}}{\sin{x}\cos{x}}$
We get:
$\displaystyle \frac{\sin^2{x}-\cos^2{x}}{\sin^2{x}+\cos^2{x}} + 1$
Recall that $\displaystyle \sin^2{x}+\cos^2{x} = 1$
Therefore:
$\displaystyle = \sin^2{x}-\cos^2{x}+1 = 2\sin^2{x}$