What is the integral value of $\frac{\tan 20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ}{\sin40^\circ}$? I have tried possibly all approaches.
I first expressed $80$ as $60+20$ and $40$ as $60-20$ and then used trig identities.I later used conditional identities expressing $\tan 20^\circ+\tan40^\circ+\tan120^\circ$ as $\tan 20^\circ \tan40^\circ \tan120^\circ$. But I really can't get to the end of it .
Please help.
 A: $$\tan20^\circ-\tan60^\circ=-\dfrac{\sin(60-20)^\circ}{\cos20^\circ\cdot\cos60^\circ}=-\dfrac{2\sin40^\circ}{\cos20^\circ}$$
$$\tan40^\circ+\tan80^\circ=\dfrac{\sin(40+80)^\circ}{\cos40^\circ\cos80^\circ}$$
Adding $(1),(2)$
$$\dfrac{\sin120^\circ}{\cos40^\circ\cos80^\circ}-\dfrac{2\sin40^\circ}{\cos20^\circ} =\dfrac{\sin120^\circ\cos20^\circ-2\sin40^\circ\cos40^\circ\cos80^\circ}
{\cos20^\circ\cos40^\circ\cos80^\circ}$$
Now $S=\sin120^\circ\cos20^\circ-2\sin40^\circ\cos40^\circ\cos80^\circ$
$2S=\sin(120+20)^\circ+\sin(120-20)^\circ-2\sin80^\circ\cos80^\circ$
$=\sin(180-40)^\circ+\sin100^\circ-\sin160^\circ$
$=\sin40^\circ+\sin80^\circ-\sin20^\circ$
$=\sin40^\circ+2\sin30^\circ\cos50^\circ$
$=2\sin40^\circ$
Formulas used :


*

*$\sin(180^\circ-A)=\sin A$

*Prosthaphaeresis Formula $:\sin C-\sin D$

*$\sin2y=2\sin y\cos y$

*$2\sin A\cos B=\sin(A+B)+\sin(A-B)$
Now use Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle?  to find the answer to be $$\dfrac1{\dfrac18}=?$$
A: Using $\tan20^\circ\cdot\tan40^\circ\cdot\tan80^\circ=\tan60^\circ$ (Proof)
$$\tan20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ$$
$$=\tan20^\circ+\tan40^\circ+\tan80^\circ-\tan20^\circ\cdot\tan40^\circ\cdot\tan80^\circ$$
$$=\tan20^\circ(1-\tan40^\circ\cdot\tan80^\circ)+\tan40^\circ+\tan80^\circ$$
$$=(1-\tan40^\circ\cdot\tan80^\circ)\left(\tan20^\circ+\dfrac{\tan40^\circ+\tan80^\circ}{1-\tan40^\circ\cdot\tan80^\circ}\right)$$
$$=\dfrac{\cos(40^\circ+80^\circ)}{\cos40^\circ\cos80^\circ}\left(\tan20^\circ+\tan(40^\circ+80^\circ)\right)$$
$$=\dfrac{\cos120^\circ}{\cos40^\circ\cos80^\circ}\cdot\dfrac{\sin(20^\circ+120^\circ)}{\cos20^\circ\cdot\cos120^\circ}$$
$$=\dfrac{\sin40^\circ}{\cos20^\circ\cos40^\circ\cos80^\circ}\text{ Using }\sin(180^\circ-A)=\sin A$$
Now use Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
A: Using the comment by  Roman83, $$\tan20^\circ+\tan80^\circ−\tan40^\circ=3\tan60^\circ$$(Proof is in Method$\#2$ here)
the numerator can be reduced to 
$$2(\tan60^\circ+\tan40^\circ)=\dfrac{2\sin(60+40)^\circ}{\cos60^\circ\cdot\cos40^\circ}$$
Using $\sin(180^\circ-A)=\sin A,\sin100^\circ=\sin80^\circ$
Using $\sin2B=2\sin B\cos B,\sin80^\circ=2\sin40^\circ\cos40^\circ$ 
and we are done!
