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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.

Blue
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Shinobi
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  • 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 – Shinobi 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 – Shinobi Jul 30 '18 at 14:46
  • See https://math.stackexchange.com/questions/455070/proving-a-fact-tan6-circ-tan42-circ-tan12-circ-tan24-cir/455573 – lab bhattacharjee Jul 30 '18 at 17:17
  • @user580093 People here are glad to see neat, clear description of the questions. So you'd better learn to type MathJax. Also, don't let us think you are imperative. Maybe you does not want to mean that but your post would be kind of commanding. Be polite. – xbh Jul 30 '18 at 18:12

2 Answers2

0

$\cos A\cos B\cos C=\cos(60^\circ-\theta).\cos(60^\circ+\theta).\cos60^\circ$

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 $

Love Invariants
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0

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.

xbh
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