Proving

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September 19th 2012, 06:45 AM
gashbell1

Proving

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

Please prove.
September 19th 2012, 06:47 AM
Prove It

Re: Proving

Quote:

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?

$\displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{1}{\cot{x} - \tan{x}} \end{align*}$
September 19th 2012, 07:30 AM
gashbell1

Re: Proving

Edited. Thanks
September 19th 2012, 07:37 AM
Prove It

Re: Proving

Quote:

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.

$\displaystyle \begin{align*} RHS &= \frac{2}{\cot{x} - \tan{x}} \\ &= \frac{2}{\frac{1}{\tan{x}} - \tan{x}} \\ &= \frac{2}{\frac{1 - \tan^2{x}}{\tan{x}}} \\ &= \frac{2\tan{x}}{1 - \tan^2{x}} \\ &= LHS \end{align*}$
September 19th 2012, 07:43 AM
gashbell1

Re: Proving

But what about left to right?
September 19th 2012, 07:44 AM
Prove It

Re: Proving

Quote:

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.
September 19th 2012, 09:00 AM
Soroban

Re: Proving

Hello, gashbell1!

Quote:

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

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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