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 
 A: 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.
A: Left side -
$\tan 70 - \tan 20$
$=  \tan(90 - 20) - \tan 20$
$ = \cot 20 - \tan 20$
$= \frac{\cos^2 20 - \sin^2 20}{\cos 20 \sin 20}$
$ = \frac{\cos 40}{\frac12 \sin 40}$
$ = 2 \cot 40$
Similarly solve for right side.
A: You can go foward with your idea.
First, note that $\tan 70°=\tan(90°-20°)=\cot20°$, so $\tan 70° \tan 20°=1$. So
$$\tan70°-\tan20°=\frac{2(\tan40°+\tan10°)}{1-\tan40°\tan10°}$$
and now you can use equivalence
$$\frac{2\tan40°+2\tan10°}{1-\tan40°\tan10°}=2\tan40°+4\tan10°\Leftrightarrow \\
2\tan40°+2\tan10°=2\tan40°+4\tan10°-\tan40°\tan10°(2\tan40°+4\tan10°)\Leftrightarrow\\
\tan10°=\tan40°\tan10°(\tan40°+2\tan10°)\Leftrightarrow 1=\tan40°(\tan40°+2\tan10°)\\
\Leftrightarrow 1-\tan^{2}40°=2\tan40°\tan10°\Leftrightarrow 1=\left(\frac{2\tan40°}{1-\tan^{2}40°}\right)\tan10°\\
\Leftrightarrow 1=\tan80°\tan 10°\Leftrightarrow 1=\cot10°\tan 10°$$
and we are done!
A: We need to prove that
$$\frac{\sin50^{\circ}}{\cos70^{\circ}\cos20^{\circ}}=\frac{2\sin50^{\circ}}{\cos40^{\circ}\cos10^{\circ}}+\frac{2\sin10^{\circ}}{\cos10^{\circ}}$$ or
$$\frac{\sin50^{\circ}\left(\frac{1}{2}\cos50^{\circ}+\frac{1}{2}\cos30^{\circ}-\cos90^{\circ}-\cos50^{\circ}\right)}{\cos70^{\circ}\cos20^{\circ}\cos40^{\circ}\cos10^{\circ}}=\frac{2\sin10^{\circ}}{\cos10^{\circ}}$$ or
$$\frac{\sin50^{\circ}\sin10^{\circ}\sin40^{\circ}}{\cos70^{\circ}\cos20^{\circ}\cos40^{\circ}\cos10^{\circ}}=\frac{2\sin10^{\circ}}{\cos10^{\circ}}$$ or
$$\sin40^{\circ}=2\cos70^{\circ}\cos20^{\circ},$$
which is obvious.
