Question

Prove the following :
$\frac{1}{\tan 3A-\tan A}-\frac{1}{\cot 3A-\cot A}=\cot 2A$

Answer 1

Given, $\frac{1}{\tan 3A-\tan A}-\frac{1}{\cot 3A-\cot A}=\cot 2A$

Changing into sin and cos, we get

$\frac{cosAcos3A}{sin3AcosA-cos3AsinA}-\frac{sin3AsinA}{cos3AsinA-sin3AcosA}$

$=\frac{cosAcos3A}{sin(3A-A)}+\frac{sin3AsinA}{sin(3A-A)}$

$=\frac{cos(3A-A)}{sin2A}=\frac{cos2A}{sin2A}$

$=cot2A$