1. ## Proving

\displaystyle \displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{2}{\cot{x} - \tan{x}} \end{align*}

2. ## Re: Proving

Originally Posted by gashbell1

2tan x 2
-------- = --------
1- tan2x cot x - tan x

please prove that L = R

??? =? R

Thanks
Is this what you're trying to prove?

\displaystyle \displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{1}{\cot{x} - \tan{x}} \end{align*}

3. ## Re: Proving

Edited. Thanks

4. ## Re: Proving

Originally Posted by gashbell1
\displaystyle \displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{2}{\cot{x} - \tan{x}} \end{align*}

\displaystyle \displaystyle \begin{align*} RHS &= \frac{2}{\cot{x} - \tan{x}} \\ &= \frac{2}{\frac{1}{\tan{x}} - \tan{x}} \\ &= \frac{2}{\frac{1 - \tan^2{x}}{\tan{x}}} \\ &= \frac{2\tan{x}}{1 - \tan^2{x}} \\ &= LHS \end{align*}

5. ## Re: Proving

But what about left to right?

6. ## Re: Proving

Originally Posted by gashbell1
But what about left to right?
That is redundant. To show two things are equal, you can go either way. It is always best to choose whichever way is easier.

7. ## Re: Proving

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 $\displaystyle \tan x$

. . . $\displaystyle \dfrac{\dfrac{2\tan x}{\tan x}}{\dfrac{1}{\tan x} - \dfrac{\tan^2x}{\tan x}} \;=\; \dfrac{2}{\cot x - \tan x}$