If tan x + tan left(x + frac{pi}{3}right) + t | vedclass

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(C) Given the equation: $	an x + 	an \left(x + \frac{\pi}{3}ight) + 	an \left(x + \frac{2\pi}{3}ight) = 3$. Using the formula $	an(A+B) = \fra

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(C) Given the equation: $	an x + 	an \left(x + \frac{\pi}{3}ight) + 	an \left(x + \frac{2\pi}{3}ight) = 3$. Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have: $	an x + \frac{	an x + \sqrt{3}}{1 - \sqrt{3} 	an x} + \frac{	an x - \sqrt{3}}{1 + \sqrt{3} 	an x} = 3$. Combining the last two terms: $	an x + \frac{(	an x + \sqrt{3})(1 + \sqrt{3} 	an x) + (	an x - \sqrt{3})(1 - \sqrt{3} 	an x)}{1 - 3 	an^2 x} = 3$. Simplifying the numerator: $(	an x + \sqrt{3} 	an^2 x + \sqrt{3} + 3 	an x) + (	an x - \sqrt{3} 	an^2 x - \sqrt{3} + 3 	an x) = 8 	an x$. So,$	an x + \frac{8 	an x}{1 - 3 	an^2 x} = 3$. $\frac{	an x(1 - 3 	an^2 x) + 8 	an x}{1 - 3 	an^2 x} = 3$. $\frac{	an x - 3 	an^3 x + 8 	an x}{1 - 3 	an^2 x} = 3$. $\frac{9 	an x - 3 	an^3 x}{1 - 3 	an^2 x} = 3$. Dividing both sides by $3$: $\frac{3 	an x - 	an^3 x}{1 - 3 	an^2 x} = 1$. Since $	an 3x = \frac{3 	an x - 	an^3 x}{1 - 3 	an^2 x}$,we get $	an 3x = 1$.

If $\tan x + \tan \left(x + \frac{\pi}{3}\right) + \tan \left(x + \frac{2\pi}{3}\right) = 3$,then $\tan 3x$ is equal to

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