If the general solution of the equation frac{ | vedclass

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(B) Given the equation $\frac{	an 3x - 1}{	an 3x + 1} = \sqrt{3}$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we can

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$\frac{1}{12}$

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(B) Given the equation $\frac{	an 3x - 1}{	an 3x + 1} = \sqrt{3}$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we can rewrite the equation as:$\frac{	an 3x - 	an(\frac{\pi}{4})}{1 + 	an 3x 	an(\frac{\pi}{4})} = \sqrt{3}$.This simplifies to $	an(3x - \frac{\pi}{4}) = 	an(\frac{\pi}{3})$.The general solution for $	an 	heta = 	an \alpha$ is $	heta = n\pi + \alpha$.So,$3x - \frac{\pi}{4} = n\pi + \frac{\pi}{3}$.$3x = n\pi + \frac{\pi}{3} + \frac{\pi}{4} = n\pi + \frac{7\pi}{12}$.$x = \frac{n\pi}{3} + \frac{7\pi}{36}$.Comparing this with $x = \frac{n\pi}{p} + \frac{7\pi}{q}$,we get $p = 3$ and $q = 36$.Therefore,$\frac{p}{q} = \frac{3}{36} = \frac{1}{12}$.

If the general solution of the equation $\frac{\tan 3x - 1}{\tan 3x + 1} = \sqrt{3}$ is $x = \frac{n\pi}{p} + \frac{7\pi}{q}$ where $n, p, q \in \mathbb{Z}$,then $\frac{p}{q}$ is

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