# Help with a trig identity

• Dec 2nd 2007, 02:07 PM
CenturionMonkey
Help with a trig identity
This identity makes no sense to me:

cot(x) - tan(x) - tan(x) + cot(x) = (4 - 2sec^2(x))/tan(x)

I did this:

LS
=2cot(x) - 2tan(x)
=2{cot(x) - tan(x)}
=2[{1/tan(x)} - tan(x)]
=2[{1-tan^2(x)}/tan(x)]
=2{sec^2(x)}/tan(x)

Where did i go wrong? I don't have a "4-" term.

ps. I used the different bracket types to differentiate from different sets of brackets. i'm sorry if it means something else in math...
• Dec 2nd 2007, 05:25 PM
Soroban
Hello, CenturionMonkey!

Quote:

$\cot x - \tan x - \tan x + \cot x \:= \:\frac{4 - 2\sec^2\!x}{\tan x}$

I did this:

$\text{LS}\:=\:2\cot x - 2\tan x \:=\:2(\cot x - \tan x) \;=\;2\left(\frac{1}{\tan x} - \tan x \right)$

. . $=\:2\left(\frac{1-\tan^2\!x}{\tan x}\right) \:=\:\frac{2\,{\color{red}\sec^2\!x}}{\tan x}$ . . . . no

We have: . $2\left(\frac{1 - \tan^2\!x}{\tan x}\right)\;=\;2\left(\frac{1-(\sec^2\!x-1)}{\tan x}\right) \;=\;2\left(\frac{2-\sec^2\!x}{\tan x}\right)$

Therefore: . $\frac{4-2\sec^2\!x}{\tan x}$

• Dec 2nd 2007, 07:14 PM
CenturionMonkey
Oh right! cause 1 + tan^2(x) = sec^2(x).

thanks a lot