# How do you prove  cotA+tan2A= cotAsec2A?

Aug 4, 2015

Probably there is a shorter and more elegant way but...

#### Explanation:

Considering that:
$\cot A = \frac{\cos A}{\sin A}$
$\tan 2 A = \frac{\sin 2 A}{\cos 2 A} = \frac{2 \sin A \cos A}{{\cos}^{2} A - {\sin}^{2} A}$
$\sec 2 A = \frac{1}{\cos 2 A} = \frac{1}{{\cos}^{2} A - {\sin}^{2} A}$
${\sin}^{2} A + {\cos}^{2} A = 1$

Using these identities into your expression you get:

$\cot A + \tan 2 A$
$= \frac{\cos A}{\sin A} + \frac{2 \sin A \cos A}{{\cos}^{2} A - {\sin}^{2} A}$
$= \frac{\cos A \left({\cos}^{2} A - {\sin}^{2} A\right) + 2 {\sin}^{2} A \cos A}{\sin A \left({\cos}^{2} A - {\sin}^{2} A\right)}$
$= \frac{\cos A \left({\cos}^{2} A + {\sin}^{2} A\right)}{\sin A \left({\cos}^{2} A - {\sin}^{2} A\right)}$
$= \cot A \sec 2 A$