If frac{tan 3theta - 1}{tan 3theta + 1} = sqr | vedclass

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(B) Given: $\frac{	an 3	heta - 1}{	an 3	heta + 1} = \sqrt{3}$ We know that $	an(\frac{\pi}{4}) = 1$,so the equation becomes: $\frac{	an 3	heta

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$\frac{n\pi}{3} + \frac{7\pi}{36}$

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(B) Given: $\frac{	an 3	heta - 1}{	an 3	heta + 1} = \sqrt{3}$ We know that $	an(\frac{\pi}{4}) = 1$,so the equation becomes: $\frac{	an 3	heta - 	an(\frac{\pi}{4})}{1 + 	an 3	heta \cdot 	an(\frac{\pi}{4})} = \sqrt{3}$ Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get: $	an(3	heta - \frac{\pi}{4}) = \sqrt{3}$ Since $	an(\frac{\pi}{3}) = \sqrt{3}$,we have: $	an(3	heta - \frac{\pi}{4}) = 	an(\frac{\pi}{3})$ The general solution for $	an x = 	an \alpha$ is $x = n\pi + \alpha$: $3	heta - \frac{\pi}{4} = n\pi + \frac{\pi}{3}$ $3	heta = n\pi + \frac{\pi}{3} + \frac{\pi}{4}$ $3	heta = n\pi + \frac{4\pi + 3\pi}{12} = n\pi + \frac{7\pi}{12}$ $	heta = \frac{n\pi}{3} + \frac{7\pi}{36}$

If $\frac{\tan 3\theta - 1}{\tan 3\theta + 1} = \sqrt{3}$,then the general value of $\theta$ is

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