Question

# If tan x + $\mathrm{tan}\left(x+\frac{\mathrm{\pi }}{3}\right)+\mathrm{tan}\left(x+\frac{2\mathrm{\pi }}{3}\right)=3$, then prove that $\frac{3\mathrm{tan}x-{\mathrm{tan}}^{3}x}{1-3{\mathrm{tan}}^{2}x}=1$.

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Solution

## $\text{Given:}\phantom{\rule{0ex}{0ex}}\mathrm{tan}x+\mathrm{tan}\left(x+\frac{\mathrm{\pi }}{3}\right)+\mathrm{tan}\left(\mathrm{x}+\frac{2\mathrm{\pi }}{3}\right)=3\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}x+\frac{\mathrm{tan}x+\mathrm{tan}\frac{\mathrm{\pi }}{3}}{1-\mathrm{tan}x\mathrm{tan}\frac{\mathrm{\pi }}{3}}+\frac{\mathrm{tan}x+\mathrm{tan}\frac{2\mathrm{\pi }}{3}}{1-\mathrm{tan}x\mathrm{tan}\frac{2\mathrm{\pi }}{3}}=3\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}x+\frac{\mathrm{tan}x+\sqrt{3}}{1-\sqrt{3}\mathrm{tan}x}+\frac{\mathrm{tan}x-\sqrt{3}}{1+\sqrt{3}\mathrm{tan}x}=3\left[\mathrm{tan}120°=-\sqrt{3}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{tan}x\left(1-3{\mathrm{tan}}^{2}x\right)+\mathrm{tan}x+\sqrt{3}+\sqrt{3}{\mathrm{tan}}^{2}x+3\mathrm{tan}x+\mathrm{tan}x-\sqrt{3}-\sqrt{3}{\mathrm{tan}}^{2}x+3\mathrm{tan}x}{1-3{\mathrm{tan}}^{2}x}=3\phantom{\rule{0ex}{0ex}}⇒\frac{9\mathrm{tan}x-3{\mathrm{tan}}^{3}x}{1-3{\mathrm{tan}}^{2}x}=3\phantom{\rule{0ex}{0ex}}⇒\frac{3\mathrm{tan}x-{\mathrm{tan}}^{3}x}{1-3{\mathrm{tan}}^{2}x}=1\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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