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

2. 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

(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,
$\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}$
$\frac{\sin^2x-\cos^2x}{\sin x\cos x}$
$\frac{\cos 2x}{(1/2)\sin 2x}$
$2\cot 2x$
$\tan x$
Unless you want me to solve for $x$?