Proving

**gashbell1** · September 19th 2012, 06:45 AM

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

Please prove.

**Prove It** · September 19th 2012, 06:47 AM

Originally Posted by gashbell1

Please prove.

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

please prove that L = R

??? =^? R

Thanks

Is this what you're trying to prove?

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

**gashbell1** · September 19th 2012, 07:30 AM

Edited. Thanks

**Prove It** · September 19th 2012, 07:37 AM

Originally Posted by gashbell1

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

Please prove.

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

**gashbell1** · September 19th 2012, 07:43 AM

But what about left to right?

**Prove It** · September 19th 2012, 07:44 AM

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.

**Soroban** · September 19th 2012, 09:00 AM

Hello, gashbell1!

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

Left-to-Right:

We have: . $\frac{2\tan x}{1 - \tan^2x}$

Divide numerator and denominator by $\tan x$

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

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