# Prove the identity. tanx-cotx/tanx+cotx=sin^2x-cos^2x Have to show the statements and the rules?

Apr 7, 2018

See below.

#### Explanation:

Identities:

$\textcolor{red}{\boldsymbol{\tan x = \sin \frac{x}{\cos} x}}$

$\textcolor{red}{\boldsymbol{\cot x = \cos \frac{x}{\sin} x}}$

Substituting these in the LHS:

$\frac{\sin \frac{x}{\cos} x - \cos \frac{x}{\sin} x}{\sin \frac{x}{\cos} x + \cos \frac{x}{\sin} x}$

Adding the terms in numerator and denominator:

((sin^2x-cos^2x)/(sinxcosx))/((sin^2x+cos^2x)/(cosxsinx)

Multiplying by $\left(\cos x \sin x\right)$

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

Identity:

$\textcolor{red}{\boldsymbol{{\sin}^{2} x + {\cos}^{2} x = 1}}$

Substituting this in the denominator:

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

${\sin}^{2} x - {\cos}^{2} x$

As required:

$L H S \equiv R H S$

Apr 7, 2018

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

$\implies \frac{\tan x - \frac{1}{\tan} x}{\tan x + \frac{1}{\tan} x}$

$\implies \frac{{\tan}^{2} x - 1}{{\tan}^{2} x + 1}$

$\implies \frac{{\tan}^{2} x - 1}{{\sec}^{2} x}$ color(white)(wwwwww $\left[\text{as } {\tan}^{2} x - {\sec}^{2} x = 1\right]$

$\implies \left({\tan}^{2} \frac{x}{\sec} ^ 2 x - \frac{1}{\sec} ^ 2 x\right)$

$\implies {\sin}^{2} x - {\cos}^{2} x$