# Proving

• Sep 19th 2012, 07:45 AM
gashbell1
Proving
\displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{2}{\cot{x} - \tan{x}} \end{align*}

• Sep 19th 2012, 07:47 AM
Prove It
Re: Proving
Quote:

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 \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{1}{\cot{x} - \tan{x}} \end{align*}
• Sep 19th 2012, 08:30 AM
gashbell1
Re: Proving
Edited. Thanks
• Sep 19th 2012, 08:37 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*}

\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*}
• Sep 19th 2012, 08:43 AM
gashbell1
Re: Proving
But what about left to right?
• Sep 19th 2012, 08:44 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.
• Sep 19th 2012, 10:00 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 $\tan x$

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