$\displaystyle \displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{2}{\cot{x} - \tan{x}} \end{align*}$
Please prove.
Hello, gashbell1!
$\displaystyle \displaystyle\frac{2\tan{x}}{1 - \tan^2{x}} \;=\; \frac{2}{\cot{x} - \tan{x}}$
Left-to-Right:
We have: .$\displaystyle \frac{2\tan x}{1 - \tan^2x}$
Divide numerator and denominator by $\displaystyle \tan x$
. . . $\displaystyle \dfrac{\dfrac{2\tan x}{\tan x}}{\dfrac{1}{\tan x} - \dfrac{\tan^2x}{\tan x}} \;=\; \dfrac{2}{\cot x - \tan x} $