Prove that $\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}$ The question is: 

Prove $$\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}.$$

I had no idea how to do it & proceeded by making RHS = $\tan60^\circ$ but this doesn't make any help. Please help me solving this.

Consider $\tan x+\tan(x+60^\circ)+\tan(x+120^\circ)$, then
\begin{align}
\tan x+\tan(x+60^\circ)+\tan(x+120^\circ)&=\tan x+\frac{\tan x+\tan60^\circ}{1-\tan x\tan60^\circ}+\frac{\tan x+\tan120^\circ}{1-\tan x\tan120^\circ}\\
&=\tan x+\frac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x}+\frac{\tan x-\sqrt{3}}{1+\sqrt{3}\tan x}\\
&=\tan x+\frac{\tan x+\sqrt{3}\tan^2 x+\sqrt{3}+3\tan x}{1-3\tan^2 x}\\
&+\frac{\tan x-\sqrt{3}\tan^2 x-\sqrt{3}+3\tan x}{1-3\tan^2 x}\\
&=\frac{\tan x(1-3\tan^2 x)}{1-3\tan^2 x}+\frac{8\tan x}{1-3\tan^2 x}\\
&=\frac{9\tan x-3\tan^33 x}{1-3\tan^2 x}\\
&=3\left(\frac{3\tan x-\tan^33 x}{1-3\tan^2 x}\right)\\
&=3\tan3x
\end{align}
See triple-angle formula of trigonometric identities. Plugging in $x=10^\circ$, we get
\begin{align}
\tan 10^\circ+\tan(10^\circ+60^\circ)+\tan(10^\circ+120^\circ)&=3\tan(3\cdot10^\circ)\\
\tan 10^\circ+\tan70^\circ+\tan130^\circ&=3\tan(30^\circ)\\
\tan 10^\circ+\tan70^\circ+\tan(180^\circ-50^\circ)&=3\cdot\frac{1}{\sqrt{3}}\\
\tan 10^\circ+\tan70^\circ-\tan50^\circ&=\sqrt{3}&\qquad\blacksquare
\end{align}
It took me ages to get the prove this identity. ᕙ(^▽^)ᕗ
