Question

Prove the identity:
tan2x = (2tan x)/(1-tan^2 x)

Answer 1

Alright, I just worked this out:

\[\frac{sen2x}{\cos2x} \rightarrow \frac{2senxcosx}{\cos^2x-sen^2x}\]

We are free to multiply both denominator and denominator for the same quantity.

\[\frac{\frac{cosx2senxcosx}{cosx}}{\frac{\cos^2x (\cos^2x - sen^2x)}{\cos^2x}}\]

Now, doing few calcs you get:

\[\frac{2tanx}{\frac{\cos^2x}{\cos^2x} - \frac{sen^2x}{\cos^2x}}\]

Which eventually goes tO:

\[\frac{2tanx}{1-\tan^2x}\]

:) Bye.