Prove that :$\tan 70 - \tan 20 = 2\tan 40 + 4\tan 10$ Prove that :$\tan 70 - \tan 20 = 2\tan 40 + 4\tan 10$
My Attempt, 
$$70-20=40+10$$
$$\tan (70-20)=\tan (40+10)$$
$$\dfrac {\tan 70 - \tan 20}{1+\tan 70. \tan 20 }=\dfrac 
{\tan 40 + \tan 10 }{1-\tan 40. \tan 10 }$$ 
How should I move on?  Please help 
Thanks 

Hint:
Repeatedly use the identity $$\tan x - \cot x = \tan x - \frac{1}{\tan x} = \frac{\tan^2 x-1}{\tan x} = \frac{-2(1-\tan^2 x)}{2\tan x} = \frac{-2}{\frac{2\tan x}{1-\tan^2 x}} = \frac{-2}{\tan 2x}$$ $$ = -2\cot 2x$$ to get, $$\tan 20 - \cot 20 = -2\cot 40 \tag{1}$$ $$2(\tan 40 - \cot 40) = -4\cot 80 \tag{2}$$ Adding $(1)$ and $(2)$ gives us, $$\tan 20 + 2\tan 40 - \cot 20 = -4\cot 80 = -4\tan 10$$ $$\Rightarrow \cot 20 = \tan 70 = \tan 20 + 2\tan 40 + 4\tan 10$$ Hope it helps.
