Question

Proofing:
sinxcosx/1-2cos^2x = 1/tanx-cotx
Help :c

Answer 1

\[
\frac{\sin x \cos x}{1-2\cos^2 x} = \frac{\sin x \cos x}{\sin^2 x + \cos^2 x-2\cos^2 x} =
\frac{\sin x \cos x}{\sin^2 x - \cos^2 x} = \\ \text{ } \\ \Large
\frac{\frac {\sin x \cos x}{\sin x \cos x}}{\frac{\sin^2 x - \cos^2 x}{\sin x \cos x}} =
\frac{1}{\tan x - \cot x}
\]

Answer 2

Thanks so much! I understand now ^^

Answer 3

You are welcome.