# How do you verify [cot^2 x - tan^2 x)/[(cot x + tan x)^2] = 2(cos^2) x - 1?

Jun 11, 2016

As shown below

#### Explanation:

LHS

$= \frac{{\cot}^{2} x - {\tan}^{2} x}{\cot x + \tan x} ^ 2$

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

$= \frac{\cot x - \tan x}{\cot x + \tan x}$

Multiplying both Numerator and denominator by $\tan x$ we get

LHS

=(tanx xx(cotx - tanx))/(tanx xx(cotx + tanx)

$= \frac{\tan x \times \cot x - {\tan}^{2} x}{\tan x \times \cot x + {\tan}^{2} x}$

$= \frac{1 - {\tan}^{2} x}{1 + {\tan}^{2} x} \text{ " "putting } \tan x \times \cot x = 1$

$= \frac{1 - {\sin}^{2} \frac{x}{\cos} ^ 2 x}{1 + {\sin}^{2} \frac{x}{\cos} ^ 2 x} \text{ " "putting } \tan x = \sin \frac{x}{\cos} x$

$= \frac{1 - {\sin}^{2} \frac{x}{\cos} ^ 2 x}{1 + {\sin}^{2} \frac{x}{\cos} ^ 2 x}$

$= \frac{{\cos}^{2} x - {\sin}^{2} x}{{\cos}^{2} x + {\sin}^{2} x}$

$= \left(\frac{{\cos}^{2} x - \left(1 - {\cos}^{2} x\right)}{{\cos}^{2} x + {\sin}^{2} x}\right)$

$= \frac{2 {\cos}^{2} x - 1}{1} = \left(2 {\cos}^{2} x - 1\right) = R H S$