# Finding the value of $\tan\theta \tan(\theta+60^\circ)+\tan\theta \tan(\theta-60^\circ)+\tan(\theta + 60^\circ) \tan(\theta-60^\circ) + 3$

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.

• Check the title, there may be brackets missing. Also please use brackets to clarify what you mean by $1/ \cos A \cos B \cos C$. – Benedict W. J. Irwin Jul 30 '18 at 14:00
• For me the expression in the title seems to simplify to $0$. – Benedict W. J. Irwin Jul 30 '18 at 14:03
• I dont think your conclusion of that expression is correct. $\cos A\cos B\cos C$ seems to be $\theta$ dependent – Love Invariants Jul 30 '18 at 14:06
• Question is edited and John yes you are right the answer is 0 – user580093 Jul 30 '18 at 14:09
• Hey friends can I know why when I mostly post any questions it have so much down votes inspite of its difficulty – user580093 Jul 30 '18 at 14:46

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 $1 + (\cos(A) \cos(B)\cos(C))^{-1}$……
$\cos A\cos B\cos C=\cos(60^\circ-\theta).\cos(60^\circ+\theta).\cos60^\circ$
Since, $\cos(60^\circ-\theta).\cos(60^\circ+\theta)={1\over2}(\cos120^\circ+\cos 2\theta)$
$\Rightarrow$ $\cos(60^\circ-\theta).\cos(60^\circ+\theta).\cos 60^\circ ={1\over2}(\cos120^\circ+\cos 2\theta).\cos60^\circ = {1\over2}({-1\over2}+\cos2\theta).1/2$