# How do you verify (csc x+sec x)/(sin x+ cos x)=cot x+tan x?

Apr 10, 2016

For an identity like this, we have to be clear with the following identities.

#### Explanation:

The reciprocal identities

$\csc \theta = \frac{1}{\sin} \theta$

$\sec \theta = \frac{1}{\cos} \theta$

$\cot \theta = \frac{1}{\tan} \theta$

The quotient identities:

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

$\cot \theta = \cos \frac{\theta}{\sin} \theta$

Applying all these identities, on both sides, we get:

$\frac{\frac{1}{\sin} x + \frac{1}{\cos} x}{\sin x + \cos x} = \cos \frac{x}{\sin} x + \sin \frac{x}{\cos} x$

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

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

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

Applying the pythagorean identity ${\sin}^{2} x + {\cos}^{2} x = 1$ on the right side, we get:

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

Hopefully this helps!