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

The correct option is B $2$

$\cot \frac{2x}{3}+\tan \frac{x}{3}=\text{cosec}\frac{kx}{3}$
Put $\frac{x}{3}=\theta$ in the above equation
$\Rightarrow \cot 2\theta +\tan \theta =\text{cosec}k\theta \Rightarrow \frac{\cos 2\theta }{\sin 2\theta }+\frac{\sin \theta }{\cos \theta }=\text{cosec}k\theta \Rightarrow \frac{\cos 2\theta +2{\sin }^2\theta }{2\sin \theta \cos \theta }=\frac{1}{\sin k\theta }$

Using $\cos 2\theta =1-2{\sin }^2\theta$, we get
$\frac{1-2{\sin }^2\theta +2{\sin }^2\theta }{2\sin \theta \cos \theta }=\frac{1}{\sin k\theta }\Rightarrow \frac{1}{\sin 2\theta }=\frac{1}{\sin k\theta }\Rightarrow k=2$