# Cot2A+tanA=?

Feb 24, 2018

$\csc 2 A$

#### Explanation:

$\cot 2 A + \tan A$

$= \frac{\cos 2 A}{\sin 2 A} + \sin \frac{A}{\cos} A$

$= \frac{\cos 2 A \cos A + \sin 2 A \sin A}{\sin 2 A \cos A}$

$= \cos \frac{2 A - A}{\sin 2 A \cos A}$

$= \cos \frac{A}{\sin 2 A \cos A}$

$= \frac{1}{\sin 2 A}$

$= \csc 2 A$

Formulae:

• $\cos C \cos D + \sin C \sin D = \cos \left(C - D\right)$
• $\cos 2 A \cos A + \sin 2 A \sin A = \cos \left(2 A - A\right)$
Feb 24, 2018

= csc 2A

#### Explanation:

Call tan A = t , and apply the trig identity
$\tan 2 A = \frac{2 \tan A}{1 - {\tan}^{2} A}$
The expression becomes:
$f \left(A\right) = \cot 2 A + \tan A = \frac{1 - {t}^{2}}{2 t} + t = \frac{\left(1 - {t}^{2}\right) + 2 {t}^{2}}{2 t}$
$f \left(A\right) = \frac{1 + {t}^{2}}{2 t} = \frac{1 + {\tan}^{2} A}{2 \tan A}$
Using trig identity, replace $\left(1 + {\tan}^{2} A\right)$ by $\left({\sec}^{2} A\right)$, we get:
$f \left(A\right) = \left({\sec}^{2} A\right) \left(\frac{2 \sin A}{\cos A}\right) = \frac{\cos A}{2 \sin A . {\cos}^{2} A}$
Note that: $\sin 2 A = 2 \sin A . \cos A$ -->
$f \left(A\right) = \frac{1}{\sin 2 A} = \csc 2 A$