Question

$\frac{1}{\tan 3A-\tan A}-\frac{1}{\cot 3A-\cot A}=k\cot 2A$, then k is equal to

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

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

LHS$=\frac{1}{\frac{sin3A}{cos3A}-\frac{sinA}{cosA}}-\frac{1}{\frac{cos3A}{sin3A}-\frac{cosA}{sinA}}$

$=\frac{cos3AcosA}{sin2A}+\frac{sin3AsinA}{sin2A}$

$=cot2A$
So, $k=1$