# Simplify [2(tanx-cotx)]/(tan^2x-cot^2x)?

Jan 15, 2017

$L H S = \frac{2 \left(\tan x - \cot x\right)}{{\tan}^{2} x - {\cot}^{2} x}$

$= \frac{2 \left(\tan x - \cot x\right)}{\left(\tan x - \cot x\right) \left(\tan x + \cot x\right)}$

$= \frac{2 \tan x}{\tan x \left(\tan x + \cot x\right)}$

$= \frac{2 \tan x}{1 + {\tan}^{2} x} = \sin 2 x = R H S$

Proved

Jan 15, 2017

$\frac{2 \left(\tan x - \cot x\right)}{{\tan}^{2} x - {\cot}^{2} x} = \sin 2 x$

#### Explanation:

Using the identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$,

we can split the term $\left({\tan}^{2} x - {\cot}^{2} x\right) = \left(\tan x + \cot x\right) \left(\tan x - \cot x\right)$ and we get

$\frac{2 \left(\tan x - \cot x\right)}{{\tan}^{2} x - {\cot}^{2} x}$

= $\frac{2 \left(\tan x - \cot x\right)}{\left(\tan x + \cot x\right) \left(\tan x - \cot x\right)}$

= $\frac{2}{\left(\tan x + \cot x\right)}$

= $\frac{2}{\left(\tan x + \cot x\right)}$

= $\frac{2}{\left(\sin \frac{x}{\cos} x + \cos \frac{x}{\sin} x\right)}$

= 2/((sin^2x+cos^2x)/(sinxcosx)

= 2/(1/(sinxcosx)

= $2 \times \frac{\sin x \cos x}{1}$

= $2 \sin x \cos x$

= $\sin 2 x$