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.
 A: 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
A: 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 $(*1)$.
A: $\tan3x=\dfrac{3t-t^3}{1-3t^2}$ where $t=\tan x$
If $\tan3x=\tan3A$
$\implies t^3-3t^2\tan3A-3t+\tan3A=0$
$\implies\sum_{r=-1}^1\tan(60^\circ r+x)=\dfrac{3\tan3A}1$
Observe that $\tan3x=\tan30^\circ$  for $x=70^\circ,-50^\circ,10^\circ$
$\implies3x=180^\circ n+30^\circ$ where $n$ is any integer
$x=60^\circ n+10^\circ;n=-1,0,1$
A: You're right, @André Nicolas:
$$\tan70° - \tan50° + \tan10° = \tan(60°+10°)-\tan(60°-10°) +\tan10°=$$ $$ =\frac{\sqrt{3}+\tan10°}{1-\sqrt{3}\cdot\tan10°}+\frac{\sqrt{3}-\tan10°}{1+\sqrt{3}\cdot\tan10°}+\tan10°= 3\frac{\tan10°\cdot(3-\tan^210°)}{1-3\cdot\tan^210°}=$$          $$=3\cdot\tan3\cdot10° =3\cdot\frac{1}{\sqrt{3}}=\sqrt{3}.$$
A: The correct question should be $\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}$. Thus
\begin{align}
\tan70^\circ - \tan50^\circ + \tan10^\circ &= \sqrt{3}\\
\tan70^\circ + \tan10^\circ - \tan50^\circ &= \sqrt{3}\\
\tan(90^\circ-20^\circ) + \frac{\sin10^\circ}{\cos10^\circ} - \tan(90^\circ-40^\circ)  &= \sqrt{3}\\
\cot(20^\circ)+ \frac{\sin10^\circ}{\cos10^\circ} -\cot(40^\circ)&= \sqrt{3}\\
\frac{\cos20^\circ}{\sin20^\circ}+ \frac{\sin10^\circ}{\cos10^\circ}-\frac{\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{\cos^210^\circ-\sin^210^\circ}{2\sin10^\circ\cos10^\circ}+ \frac{2\sin10^\circ\sin10^\circ}{2\sin10^\circ\cos10^\circ}-\frac{\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{\cos^210^\circ+\sin^210^\circ}{2\sin10^\circ\cos10^\circ}-\frac{\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{1}{\sin20^\circ}-\frac{\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{2\cos20^\circ}{2\sin20^\circ\cos20^\circ}-\frac{\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{2\cos20^\circ-\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{2\cos(60^\circ-40^\circ)-\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{2\cos60^\circ\cos40^\circ+2\sin60^\circ\sin40^\circ-\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{2\cdot\frac12\cdot\cos40^\circ+2\cdot\frac12\sqrt3\cdot\sin40^\circ-\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\frac{\cos40^\circ+\sqrt3 \sin40^\circ-\cos40^\circ}{\sin40^\circ}&= \sqrt{3}\\
\end{align}

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
