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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

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    $a/b = c/d$ implies $ad = bc$ might help. – 2017-02-16
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    Also observe that $$\tan 70^{\circ}\cdot\tan 20^{\circ}=\frac{\sin 20^{\circ}}{\cos 20^{\circ}}\cdot\frac{\sin 70^{\circ}}{\cos 70^{\circ}}=1$$ Since $\sin 20^{\circ}=\cos 70^{\circ}$ and $\sin 70^{\circ}=\cos 20^{\circ}$ – 2017-02-16
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    @ Student, but that didn't give the final result – 2017-02-16
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    @ Ángel Mario Gallegos, how did that help? Please elaborate – 2017-02-16

4 Answers 4

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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.

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    @ Rohan, could you please elaborate? – 2017-02-16
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    @ÉvaristeGalois Elaborate where? It is quite straightforward. Just use some identities and we are done. – 2017-02-16
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    @ Rohan, after this 'Adding...' – 2017-02-16
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    @ÉvaristeGalois Add $(1)$ and $(2)$ to get, $$\tan 20 - \cot 20 + 2\tan 40 - 2\cot 40 = -2\cot 40 -4\cot 80$$ $$\Rightarrow \tan 20 + 2\tan 40 - \cot 20 =-4\cot 80$$ Now observe that $\cot 80 = \tan(90-80) = \tan 10$ and that $\cot 20 = \tan(90-20) = \tan 70$ because $\cot x = \tan (90-x)$ where all measures are in degrees. – 2017-02-16
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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.

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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!

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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.