On the proof $\tan 70°-\tan 20° -2 \tan 40°=4\tan 10°$

\begin{align*} \tan 70-\tan 20 & =\tan 70-\cot 70\\ & =\tan 70-\frac{1}{\tan 70}\\ & = \frac{\tan^2 70-1}{\tan 70}=-\frac{1-\tan^2 70}{\tan 70}\\ & = -\frac{2(1-\tan^2 70)}{2\tan 70}=-\frac{2}{\frac{2\tan 70}{1-\tan^2 70}}\\ & = \frac{-2}{\tan 140}= \frac{-2}{-\tan 40}= \frac{2}{\tan 40}=2\cot 40 \end{align*} Now, $$ \tan 70-\tt-2\tan 40 =2\cot 40-2\tan 40=-2(\tan 40-\cot 40) $$ \begin{align*} -2(\tan 40-\cot 40)= -2\left( \tan 40-\frac{1}{\tan 40} \right) \end{align*} Now do the same as above to get the result. You will get $ 4\cot 80=4\tan 10. $

$\tan 70 = \tan(90 - 20) = \cot 20$. So

$$\tan 70 - \tan 20 = \cot 20 - \tan 20= \frac{\cos^2 20 - \sin^2 20}{\cos 20 \sin 20} = \frac{\cos 40}{\frac12 \sin 40} = 2 \cot 40$$

Repeating the same procedure, we get that the LHS is:

$$2(\cot 40 - \tan 40) = \cdots = 2 (2 \cot 80) = 4\cot (90 - 10) = 4 \tan 10$$