Trigonometry Identities

**Giggly2** · April 29th 2009, 05:55 PM

Hey Im having trouble with these identities i was wondering if anyone else could figure them out??

cosX/cosX-sinX = 1/1-tanX

2tanX/tan(squared) -1 = 2/tanX-cotX

tanXsinX/tanX-sinX = sinX/1-cosX

(tanX + cotX)(secX + cosX) = secXtanX

**Soroban** · April 29th 2009, 06:39 PM

Hello, Giggly2!

There is a typo in the last one . . .

$\frac{\cos x}{\cos x-\sin x} \:=\:\frac{1}{1-\tan x}$

On the left side, divide top and bottom by $\cos x$

. . $\frac{\dfrac{\cos x}{\cos x}}{\dfrac{\cos x}{\cos x} - \dfrac{\sin x}{\cos x}} \;=\;\frac{1}{1 - \tan x}$

$\frac{2\tan x}{\tan^2\!x -1} \:=\: \frac{2}{\tan x-\cot x}$

On the left, divide top and bottom by $\tan x$

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

$\frac{\tan x\sin x}{\tan x-\sin x} \:=\: \frac{\sin x}{1-\cos x}$

We have: . $\frac{\dfrac{\sin x}{\cos x}\cdot\sin x}{\dfrac{\sin x}{\cos x} - \sin x}$

Multiply by $\frac{\cos x}{\cos x}\!:\;\;\frac{\cos x\left(\dfrac{\sin^2\!x}{\cos x}\right)}{\cos x\left(\dfrac{\sin x}{\cos x} - \sin x\right)} \;=\;\frac{\sin^2\!x}{\sin x -\sin x\cos x}$

Factor and reduce: . $\frac{\sin^2\!x}{\sin x(1 - \cos x)} \;=\;\frac{\sin x}{1 - \cos x}$

$(\tan x + \cot x)(\sec x \;{\color{red}-}\; \cos x) \:=\: \sec x\tan x$

The left side is: . $\left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right)\cdot\left(\frac{1}{\cos x} - \cos x\right)$

. . $= \;\frac{\overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}}}{\sin x\cos x} \cdot \frac{\overbrace{1-\cos^2\!x}^{\text{This is }\sin^2\!x}}{\cos x} \;=\;\frac{1}{\sin x\cos x}\cdot\frac{\sin^2\!x}{\cos x}$

. . $= \;\frac{\sin x}{\cos^2\!x} \;=\; \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} \;=\;\sec x \tan x$

**Giggly2** · April 29th 2009, 07:20 PM

Thank you soooo much, that was so helpful

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