Question

If $tanx+tan(x+\frac{\pi }{3})+tan(x+\frac{2\pi }{3})=3,$ then prove that $\frac{3tanx-ta{n}^{3}x}{1-3ta{n}^{2}x}=1$

Answer 1

We have,
$tanx+tan(x+\frac{\pi }{3})+tan(x+\frac{2\pi }{3})=3$
$\Rightarrow tanx+\left[\frac{tanx+tan\frac{\pi }{3}}{1-tanx\times tan\frac{\pi }{3}}\right]+\left[\frac{tanx+tan\left(\frac{2\pi }{3}\right)}{1-tanx\times tan\frac{2\pi }{3}}\right]=3$
$\Rightarrow tanx+\left[\frac{tanx+\sqrt{3}}{1-\sqrt{3}tanx}\right]+\left[\frac{tanx+tan(\frac{\pi }{2}+\frac{\pi }{3})}{1-tanxtan(\frac{\pi }{2}+\frac{\pi }{3})}\right]=3$
$tanx+\frac{tanx+\sqrt{3}}{1-\sqrt{3}tanx}+\frac{tanx-\sqrt{3}}{1+\sqrt{3}tanx}=3$
$\Rightarrow tanx+\frac{tanx+\sqrt{3}}{1-\sqrt{3}tanx}+\frac{tanx-\sqrt{3}}{1+\sqrt{3}tanx}=3$
$\Rightarrow tanx+\frac{(tanx+\sqrt{3})(1+\sqrt{3}tanx)+(tanx-\sqrt{3})(1-\sqrt{3}tanx)}{(1-\sqrt{3}tanx)(1+\sqrt{3}tanx)}=3$
$\Rightarrow tanx+\frac{tanx+\sqrt{3}ta{n}^{2}x+\sqrt{3}+3tanx+tanx-\sqrt{3}ta{n}^{2}x-\sqrt{3}+3tanx}{1-(\sqrt{3}tanx{)}^2}=3$
$\Rightarrow tanx+\frac{8tanx}{1-3ta{n}^{2}x}=3\Rightarrow \frac{tanx(1-3ta{n}^{2}x)+8tanx}{1-3ta{n}^{2}x}=3$
$\Rightarrow \frac{tanx-3ta{n}^{3}x+8tanx}{1-3ta{n}^{2}x}=3\Rightarrow \frac{9tanx-3ta{n}^{3}x}{1-3ta{n}^{2}x}=3$
$\Rightarrow \frac{3(3tanx-ta{n}^{2}x)}{1-3ta{n}^{2}x}\Rightarrow \frac{3tanx-ta{n}^{3}x}{1-3ta{n}^{2}x}=1$
Hence proved.