(tan/theta) - ((sin/theta)(cos/theta)) / (sin^2/theta) = (tan/theta)

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- Mar 22nd 2006, 08:27 AMcheerleader4life4uPlease help Trig FUnctions!!!!!
(tan/theta) - ((sin/theta)(cos/theta)) / (sin^2/theta) = (tan/theta)

- Mar 22nd 2006, 09:05 AMTD!
Bring everything to one side, convert it all to either sin and cos or tan and cot to simplify.

Give it a shot, if it doesn't work out we'll help further :) - Mar 22nd 2006, 01:26 PMThePerfectHackerQuote:

Originally Posted by**cheerleader4life4u**

$\displaystyle \tan x-\frac{\sin x\cos x}{\sin^2x}$

First you can cancel the sines, to get,

$\displaystyle \tan x-\frac{\cos x}{\sin x}$

Next, express tangent through sine and cosine,

$\displaystyle \frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}$

Add fractions,

$\displaystyle \frac{\sin^2x-\cos^2x}{\sin x\cos x}$

Recognize the double-angle identities,

$\displaystyle \frac{\cos 2x}{(1/2)\sin 2x}$

Thus,

$\displaystyle 2\cot 2x$

Your claim that it is equivalent to

$\displaystyle \tan x$

is dreadfully wrong.

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Unless you want me to solve for $\displaystyle x$?