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

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

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 $(*1)$.

$\tan3x=\dfrac{3t-t^3}{1-3t^2}$ 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

$x=60^\circ n+10^\circ;n=-1,0,1$

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

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}

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