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

Jun 7, 2016

Recall the reciprocal identities:

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

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

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

Also, the quotient identities will be helpful

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

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

Now, simplify both sides:

$\left(\frac{1}{\cos} \theta + \frac{1}{\sin} \theta\right) \left(\cos \theta - \sin \theta\right) = \cos \frac{\theta}{\sin} \theta - \sin \frac{\theta}{\cos} \theta$

(sin theta + costheta)/(costhetasintheta)xx (costheta - sintheta) = (cos^2theta - sin^2theta)/(costhetasintheta

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

$\frac{{\sin}^{2} \theta - {\cos}^{2} \theta}{\cos \theta \sin \theta} = \frac{{\cos}^{2} \theta - {\sin}^{2} \theta}{\cos \theta \sin \theta}$

Identity proved!!