Question: 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.

Answer: 
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](http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae). 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
If you know following cotangent identity,
$$n \cot(n\theta) = \sum\_{k=0}^{n-1}\cot\left(\theta+\frac{k\pi}{n}\right)
\quad\forall n \in \mathbb{Z}\_{+}\tag{\*1}$$
What you need to show is pretty simple,
$$\tan10^\circ + \tan 70^\circ - \tan 50^\circ = \cot 80^\circ + \cot 20^\circ + \cot( -40^\circ)\\
= \sum\_{k=0}^{2}\cot\left(-40^\circ + \frac{k}{3} \times 180^\circ\right)
= 3\cot(3 \times -40^\circ) = 3 \cot 60^\circ = \frac{3}{\sqrt{3}} = \sqrt{3}.
$$
Since I didn't find a proof of the cotangent identity online. I will give a proof here.
For any $\theta$ and integer $n > 0$, notice

$$\begin{align}
\prod\_{k=0}^{n-1}\sin\left(\theta + \frac{k\pi}{n}\right)
&= \prod\_{k=0}^{n-1}\left[\frac{
e^{i\left(\theta + \frac{k\pi}{n}\right)} - e^{-i\left(\theta + \frac{k\pi}{n}\right)}
}{2i}\right]
= \frac{e^{-i\left(n\theta - \sum\_{k=0}^n \frac{k\pi}{n}\right)}}{(2i)^n}
\prod\_{k=0}^{n-1}\left(e^{2i\theta}-e^{-i\frac{2k\pi}{n}}\right)\\
&= \frac{e^{-in\theta}}{2^ni}\left(e^{2in\theta} - 1\right)
= 2^{1-n} \sin(n\theta)
\end{align}
$$
Taking logarithm and differentiate on both sides immediately give you identity $(\*1)$.