Please help Trig FUnctions!!!!!

cheerleader4life4u · Mar 22nd 2006, 09:27 AM

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

**TD!** · Mar 22nd 2006, 10:05 AM

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

**ThePerfectHacker** · Mar 22nd 2006, 02:26 PM

Originally Posted by cheerleader4life4u

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

I understand this is as,
$\tan x-\frac{\sin x\cos x}{\sin^2x}$
First you can cancel the sines, to get,
$\tan x-\frac{\cos x}{\sin x}$
Next, express tangent through sine and cosine,
$\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}$
Add fractions,
$\frac{\sin^2x-\cos^2x}{\sin x\cos x}$
Recognize the double-angle identities,
$\frac{\cos 2x}{(1/2)\sin 2x}$
Thus,
$2\cot 2x$
Your claim that it is equivalent to
$\tan x$
is dreadfully wrong.
---
Unless you want me to solve for $x$ ?

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