# How do you verify (cot theta - tan theta)/(cot theta + tan theta) = cos 2 theta?

Taking LHS as follows

$L H S = \setminus \frac{\setminus \cot \setminus \theta - \setminus \tan \setminus \theta}{\setminus \cot \setminus \theta + \setminus \tan \setminus \theta}$

$= \setminus \frac{\setminus \cos \setminus \frac{\theta}{\setminus} \sin \setminus \theta - \setminus \sin \setminus \frac{\theta}{\setminus} \cos \setminus \theta}{\setminus \cos \setminus \frac{\theta}{\setminus} \sin \setminus \theta + \setminus \sin \setminus \frac{\theta}{\setminus} \cos \setminus \theta}$

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

$= \setminus \frac{\setminus {\cos}^{2} \setminus \theta - \left(1 - \setminus {\cos}^{2} \setminus \theta\right)}{1}$

$= \setminus {\cos}^{2} \setminus \theta - 1 + \setminus {\cos}^{2} \setminus \theta$

$= 2 \setminus {\cos}^{2} \setminus \theta - 1$

$= \setminus \cos 2 \setminus \theta$

$= R H S$

proved.

Jun 30, 2018

Shown below

#### Explanation:

Let $t = \tan \left(\frac{a}{2}\right)$

$\sin a = \frac{2 t}{1 + {t}^{2}}$

$\cos a = \frac{1 - {t}^{2}}{1 + {t}^{2}}$

$\implies \tan a = \frac{2 t}{1 - {t}^{2}}$

$L H S :$

$= \frac{\frac{1 - {t}^{2}}{2 t} - \frac{2 t}{1 - {t}^{2}}}{\frac{1 - {t}^{2}}{2 t} + \frac{2 t}{1 - {t}^{2}}}$

$= \frac{1 - {t}^{2}}{1 + {t}^{2}}$

$= \cos a$

$= R H S$

$\implies \frac{\cot \left(\frac{a}{2}\right) - \tan \left(\frac{a}{2}\right)}{\cot \left(\frac{a}{2}\right) + \tan \left(\frac{a}{2}\right)} = \cos a$

let $\theta = \frac{a}{2} \implies 2 \theta = a$

Hence $\frac{\cot \left(\theta\right) - \tan \left(\theta\right)}{\cot \left(\theta\right) + \tan \left(\theta\right)} = \cos 2 \theta$

This idea of using $t = \tan \left(\frac{\theta}{2}\right)$ can be used for a great amount of problems similar to this one, and even for integrals like $\int \frac{1}{1 + \sin \theta} d \theta$ where its called weirstrass substitution