How do you factor and simplify tan^2x-cot^2x?

Nov 29, 2016

$- \frac{4 \left(\sin x - \cos x\right) \left(\sin x + \cos x\right)}{{\sin}^{2} 2 x}$ or
-4cot 2x.csc 2x

Explanation:

${\tan}^{2} x - {\cot}^{2} x = {\sin}^{2} \frac{x}{{\cos}^{2} x} - {\cos}^{2} \frac{x}{{\sin}^{2} x} =$
$= \frac{{\sin}^{4} x - {\cos}^{4} x}{{\sin}^{2} x . {\cos}^{2} x} =$
Since:
$\left(a\right) = \left({\sin}^{4} x - {\cos}^{4} x\right) = \left({\sin}^{2} x - {\cos}^{2} x\right) \left({\sin}^{2} x + {\cos}^{2} x\right) =$
$= \left(\sin x - \cos x\right) \left(\sin x + \cos x\right)$ , and
$\left(b\right) = {\sin}^{2} x . {\cos}^{2} x = \left(\frac{1}{4}\right) {\sin}^{2} 2 x$,
There for:
${\tan}^{2} x - {\cot}^{2} x = \frac{\left(a\right)}{\left(b\right)} = \frac{4 \left(\sin x - \cos x\right) \left(\sin x + \cos x\right)}{{\sin}^{2} 2 x}$

There is another answer for simplification:
Since (sin^2 x - cos^2 x) = - cos 2x, then,
$\frac{\left(a\right)}{\left(b\right)} = - \frac{4 \cos 2 x}{{\sin}^{2} 2 x} = - \left(4 \cot 2 x\right) \left(\frac{1}{\sin 2 x}\right) =$
$- 4 \cot 2 x . \csc 2 x$ #