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

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

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$	an 3x = 1$

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(C) Given the expression: $	an x + 	an \left( \frac{\pi}{3} + x ight) + 	an \left( \frac{2\pi}{3} + x ight) = 3$. Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have: $	an \left( \frac{\pi}{3} + x ight) = \frac{\sqrt{3} + 	an x}{1 - \sqrt{3} 	an x}$ and $	an \left( \frac{2\pi}{3} + x ight) = \frac{-\sqrt{3} + 	an x}{1 + \sqrt{3} 	an x}$. Summing these: $	an x + \frac{\sqrt{3} + 	an x}{1 - \sqrt{3} 	an x} + \frac{	an x - \sqrt{3}}{1 + \sqrt{3} 	an x} = 	an x + \frac{(\sqrt{3} + 	an x)(1 + \sqrt{3} 	an x) + (	an x - \sqrt{3})(1 - \sqrt{3} 	an x)}{1 - 3 	an^2 x}$. $= 	an x + \frac{\sqrt{3} + 3 	an x + 	an x + \sqrt{3} 	an^2 x + 	an x - \sqrt{3} 	an^2 x - \sqrt{3} + 3 	an x}{1 - 3 	an^2 x} = 	an x + \frac{8 	an x}{1 - 3 	an^2 x}$. $= \frac{	an x - 3 	an^3 x + 8 	an x}{1 - 3 	an^2 x} = \frac{9 	an x - 3 	an^3 x}{1 - 3 	an^2 x} = 3 \left( \frac{3 	an x - 	an^3 x}{1 - 3 	an^2 x} ight) = 3 	an 3x$. Given $3 	an 3x = 3$,we get $	an 3x = 1$.

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

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