If cot frac{2x}{3} + tan frac{x}{3} = operato | vedclass

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(B) Given, $\cot \frac{2x}{3} + 	an \frac{x}{3} = \operatorname{cosec} \frac{kx}{3}$.Let $	heta = \frac{x}{3}$.Then, $\cot 2	heta + 	an 	heta = \

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$2$

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(B) Given, $\cot \frac{2x}{3} + 	an \frac{x}{3} = \operatorname{cosec} \frac{kx}{3}$.Let $	heta = \frac{x}{3}$.Then, $\cot 2	heta + 	an 	heta = \operatorname{cosec} k	heta$.Using the identities $\cot 2	heta = \frac{\cos 2	heta}{\sin 2	heta}$ and $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$:$\frac{\cos 2	heta}{\sin 2	heta} + \frac{\sin 	heta}{\cos 	heta} = \operatorname{cosec} k	heta$.$\frac{\cos 2	heta \cos 	heta + \sin 2	heta \sin 	heta}{\sin 2	heta \cos 	heta} = \operatorname{cosec} k	heta$.Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$:$\frac{\cos(2	heta - 	heta)}{\sin 2	heta \cos 	heta} = \operatorname{cosec} k	heta$.$\frac{\cos 	heta}{\sin 2	heta \cos 	heta} = \operatorname{cosec} k	heta$.$\frac{1}{\sin 2	heta} = \operatorname{cosec} k	heta$.$\operatorname{cosec} 2	heta = \operatorname{cosec} k	heta$.Therefore, $k = 2$.

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

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