# How do you verify the identity (cos^2theta-sin^2theta)=(cottheta-tantheta)/(tantheta+cottheta)?

Sep 2, 2016

See explanation...

#### Explanation:

Use:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

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

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

Hence:

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

$\textcolor{w h i t e}{{\cos}^{2} \theta - {\sin}^{2} \theta} = \frac{\left({\cos}^{2} \theta - {\sin}^{2} \theta\right) \div \left(\cos \theta \sin \theta\right)}{\left({\sin}^{2} \theta + {\cos}^{2} \theta\right) \div \left(\cos \theta \sin \theta\right)}$

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

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