Question

If $\cot \frac{2x}{3}+\tan \frac{x}{3}=\text{cosec}\frac{kx}{3}$, then the value of $k$ is

• A. $1$
• B. $2$
• C. $3$
• D. $-1$

Answer 1

Answer: (B)

$2$

Given,

$\cot \frac{2x}{3}+\tan \frac{x}{3}=\text{cosec}\frac{kx}{3}$

Let, $\theta =\frac{x}{3}$, then
$\therefore \cot 2\theta +\tan \theta =\text{cosec}k\theta$

$\Rightarrow \frac{\cos 2\theta }{\sin 2\theta }+\frac{\sin \theta }{\cos \theta }=\text{cosec}k\theta$
$\Rightarrow \frac{2{\cos }^2\theta -1}{2\sin \theta \cos \theta }+\frac{\sin \theta }{\cos \theta }=\text{cosec}k\theta$
$\Rightarrow \frac{2{\cos }^2\theta -1+2{\sin }^2\theta }{2\sin \theta \cos \theta }=\text{cosec}k\theta$
$\Rightarrow \frac{1}{2\sin \theta \cos \theta }=\text{cosec}k\theta$
$\Rightarrow \text{cosec}2\theta =\text{cosec}k\theta$
$\therefore k=2$