Proving a tangent value equality

Question

Show that $\tan50 \tan60 \tan70 = \tan80$.

I have used compound angle formula for tan to bring $\tan 10$ into it as $\frac{1}{\tan 10}=\tan 80$, but I can't seem to get it to come out.

Answer 1

In triangle we have $$\tan \alpha +\tan \beta +\tan \gamma = \tan \alpha \tan \beta \tan \gamma$$ so we have to prove:

$$\tan 50 +\tan 60 +\tan 70 = \tan 80$$

or

$$\tan 50 +\tan 60 =\tan 80 - \tan 70$$

or $$ {\sin 110\over \cos 50 \cos 60} = {1\over \cos 70}$$

so we are left to prove $$ 2\sin 70 \cos 70 = \cos 50 $$ or $$ 2\cos 20 \sin 20 = \sin 40$$ which is true.

Answer 2

We can prove $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$ (Proof)

$x=10^\circ\implies$

$$\tan10^\circ\tan(60^\circ-10^\circ)\tan(60^\circ+10^\circ)=\tan3(10^\circ)$$

Now use $\tan(90^\circ-y)=\dfrac1{\tan y}$ for $y=30^\circ,10^\circ$