<s>[INST] 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.
 [/INST] 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 $(*1)$.
</s>