# Prove that (tanx-cotx)/(tanx+cotx)=sin^2x-cos^2x?

Nov 26, 2016

#### Explanation:

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

= $\frac{\sin \frac{x}{\cos} x - \cos \frac{x}{\sin} x}{\sin \frac{x}{\cos} x + \cos \frac{x}{\sin} x}$ - (converting tan ad cot ratios into sin and cos ratios)

= $\frac{\frac{{\sin}^{2} x - {\cos}^{2} x}{\sin x \cos x}}{\frac{{\sin}^{2} x + {\cos}^{2} x}{\sin x \cos x}}$ - (adding as in fractions)

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

= $\frac{{\sin}^{2} x - {\cos}^{2} x}{\cancel{\sin x \cos x}} \times \frac{\cancel{\sin x \cos x}}{1}$ - (as ${\sin}^{2} x + {\cos}^{2} x = 1$ and cancelling like terms in numerator and denominator)

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