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: 
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 $(\*1)$.
You're right, @André Nicolas:
$$\tan70° - \tan50° + \tan10° = \tan(60°+10°)-\tan(60°-10°) +\tan10°=$$ $$ =\frac{\sqrt{3}+\tan10°}{1-\sqrt{3}\cdot\tan10°}+\frac{\sqrt{3}-\tan10°}{1+\sqrt{3}\cdot\tan10°}+\tan10°= 3\frac{\tan10°\cdot(3-\tan^210°)}{1-3\cdot\tan^210°}=$$ $$=3\cdot\tan3\cdot10° =3\cdot\frac{1}{\sqrt{3}}=\sqrt{3}.$$