# How do you prove the identity (sinx-cosx)/cos^2x=(tan^2x-1)/(sinx+cosx)?

Sep 30, 2016

See explanation...

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Use this with $a = \sin x$ and $b = \cos x$ to find:

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

color(white)((sin x - cos x)/(cos^2 x)) = (sin^2 x - cos^2 x)/((cos^2 x)(sin x + cos x)

$\textcolor{w h i t e}{\frac{\sin x - \cos x}{{\cos}^{2} x}} = \frac{\frac{{\sin}^{2} x}{{\cos}^{2} x} - \frac{{\cos}^{2} x}{{\cos}^{2} x}}{\sin x + \cos x}$

$\textcolor{w h i t e}{\frac{\sin x - \cos x}{{\cos}^{2} x}} = \frac{{\tan}^{2} x - 1}{\sin x + \cos x}$