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

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Question

(B) Given the equation: $\cot \frac{2x}{3} + 	an \frac{x}{3} = \csc \frac{kx}{3}$.Using the identities $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}

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$2 - \pi$

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(B) Given the equation: $\cot \frac{2x}{3} + 	an \frac{x}{3} = \csc \frac{kx}{3}$.Using the identities $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$ and $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we have:$\frac{\cos(2x/3)}{\sin(2x/3)} + \frac{\sin(x/3)}{\cos(x/3)} = \frac{\cos(2x/3)\cos(x/3) + \sin(2x/3)\sin(x/3)}{\sin(2x/3)\cos(x/3)}$.Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$,the numerator becomes $\cos(2x/3 - x/3) = \cos(x/3)$.So,the expression is $\frac{\cos(x/3)}{\sin(2x/3)\cos(x/3)} = \frac{1}{\sin(2x/3)} = \csc \frac{2x}{3}$.Comparing this with $\csc \frac{kx}{3}$,we get $k = 2$.We need to find $	an^{-1}(	an 2)$.Since $2$ radians is in the second quadrant $(\pi/2 < 2 < \pi)$,the principal value of $	an^{-1}(	an 2)$ is $2 - \pi$.

If $\cot \frac{2x}{3} + \tan \frac{x}{3} = \csc \frac{kx}{3}$,then the value of $\tan^{-1}(\tan k)$ equals:

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