Find the value of $$\tan\theta \tan(\theta+60^\circ)+\tan\theta \tan(\theta-60^\circ)+\tan(\theta + 60^\circ) \tan(\theta-60^\circ) + 3$$ (The answer is $0$.)
My try: Let $\theta$ be $A$, $60^\circ -\theta$ be $B$, and $60^\circ + \theta$ be $C$. I simplified the result and got the expression $$1 + 1/\cos A\cos B\cos C$$ but after that I can't simplify it.
On
$\cos A\cos B\cos C=\cos(60^\circ-\theta).\cos(60^\circ+\theta).\cos60^\circ$
Since, $\cos(60^\circ-\theta).\cos(60^\circ+\theta)={1\over2}(\cos120^\circ+\cos 2\theta)$
$\Rightarrow$ $\cos(60^\circ-\theta).\cos(60^\circ+\theta).\cos 60^\circ ={1\over2}(\cos120^\circ+\cos 2\theta).\cos60^\circ = {1\over2}({-1\over2}+\cos2\theta).1/2 $
Recall that $$ \tan (A - B) = \frac {\tan (A )-\tan (B)} { 1+\tan (A) \tan (B)}, $$ which means $$ 1 + \tan (A) \tan (B) = \frac {\tan (A) - \tan(B)} {\tan (A-B)}. $$ Thus the answer is $$ \frac {\tan (\theta + 60^\circ) - \tan (\theta ) } {\sqrt 3} + \frac {\tan (\theta ) - \tan(\theta - 60^\circ)} {\sqrt 3} - \frac {\tan (\theta + 60^\circ) - \tan (\theta - 60^\circ)} {\sqrt 3} = 0. $$
I have no idea how to transform this to $1 + (\cos(A) \cos(B)\cos(C))^{-1}$……
P.S. If the formula involving the tangent of sums seems unfamiliar, you may prove it by those formulae about sines and cosines.