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If $$(\tan\alpha)^2 (\tan\beta)^2 + (\tan\beta)^2 (\tan\gamma)^2 +(\tan\alpha)^2 (\tan\gamma)^2 + 2(\tan\alpha)^2(\tan\beta)^2 (\tan\gamma)^2 = 1 $$ (where $\alpha ,\beta ,\gamma$ are as per domain of $\tan x$ ). Find the value of $$\cos2\alpha +\cos2\beta+\cos2\gamma$$

I had got it by arbitrarily choosing some specific angles so that the given condition is satisfied. But it's a negative approach. So I am looking for more of a general approach....

So What I did is I expanded $\cos2\alpha$ in $\tan\alpha$ form and cross multiplied everything but didn't get anything.

Thanks for any help in advance.

ACB
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Krishna Patil
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    Do the three angles make a triangle? If so you should say so. Consider using $\cos 2x=\frac{1-\tan^2 x}{1+\tan^2 x}$. – Blitzer May 22 '23 at 11:05
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    No actually those angles aren't angles of triangle – Krishna Patil May 22 '23 at 11:25
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    I tried using the formula you mentioned but nothing is happening.... – Krishna Patil May 22 '23 at 11:27
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    I had got it my arbitrarily choose some specific angles so that the given condition is satisfied.But it's a negative approach .So I am looking for more of a general approach.... – Krishna Patil May 22 '23 at 11:55
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    Start with $\cos(2\alpha) = 1-2\sin^2 \alpha$, $\cos(2 \beta) = 1-2\sin^2\beta$, $\cos(2\gamma) = 1-2\sin^2 \gamma$. Adding them gives that it's enough to know $\sin^2 \alpha+\sin^2 \beta + \sin^2 \gamma$. Now see either the last post here, or here or here. The approaches are general, they don't follow by guessing $\alpha,\beta,\gamma$. The last post is a dupe candidate, IMO. – Sarvesh Ravichandran Iyer May 22 '23 at 12:48

2 Answers2

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Starting with $\cos(2a)+\cos(2b)+\cos(2c)$, and use the tangent version of the double angle formula (for brevity's sake, let $A=(\tan(a))^2,B=(\tan(b))^2,C=(\tan(c))^2$) $$\cos(2a)+\cos(2b)+\cos(2c)=\frac{1-A}{1+A}+\frac{1-B}{1+B}+\frac{1-C}{1+C}$$

Then, condense the RHS into a common denominator, and apply algebra: $$\frac{1-A}{1+A}+\frac{1-B}{1+B}+\frac{1-C}{1+C}=\frac{(1-A)(1+B)(1+C)}{(1+A)(1+B)(1+C)}+\frac{(1+A)(1-B)(1+C)}{(1+A)(1+B)(1+C)}+\frac{(1+A)(1+B)(1-C)}{(1+A)(1+B)(1+C)}=\frac{1-A+B+C-AB-AC+BC-ABC}{1+A+B+C+AB+AC+BC+ABC}+\frac{1+A-B+C-AB+AC-BC-ABC}{1+A+B+C+AB+AC+BC+ABC}+\frac{1+A+B-C+AB-AC-BC-ABC}{1+A+B+C+AB+AC+BC+ABC}=\frac{3+A+B+C-AB-AC-BC-3ABC}{1+A+B+C+AB+AC+BC+ABC}$$

Now, we were given $AB+AC+BC+2ABC=1$. We can manipulate the above to use this substitution to simplify: $$\frac{3+A+B+C-AB-AC-BC-3ABC}{1+A+B+C+AB+AC+BC+ABC}=\frac{3+A+B+C-AB-AC-BC-ABC-2ABC}{1+A+B+C+AB+AC+BC+ABC+ABC-ABC}=\frac{3+A+B+C-(AB+AC+BC+2ABC)-ABC}{1+A+B+C+(AB+AC+BC+2ABC)-ABC}=\frac{3+A+B+C-1-ABC}{1+A+B+C+1-ABC}=\frac{2+A+B+C-ABC}{2+A+B+C-ABC}=1$$

Moko19
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By $\cos 2x=\frac{1-\tan^2 x}{1+\tan^2 x}$, and let $a^2=\tan^2\alpha, b^2=\tan^2\beta, c^2= \tan^2\gamma$, we have:

\begin{align} \cos2\alpha +\cos2\beta+\cos2\gamma &=\frac{1-a^2}{1+a^2}+\frac{1-b^2}{1+b^2}+\frac{1-c^2}{1+c^2}\\ &=-3+2(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2})\\ \end{align} For $\displaystyle\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}$, we have: \begin{align} \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2} &=\frac{3+2(a^2+b^2+c^2)+a^2b^2+a^2c^2+b^2c^2+a^2b^2c^2}{(1+a^2)(1+b^2)(1+c^2)} \end{align}

By the given condition, we have: \begin{align} (1+a^2)(1+b^2)(1+c^2)&=1+a^2+b^2+c^2+a^2b^2+a^2c^2+b^2c^2+a^2b^2c^2\\ &=2+a^2+b^2+c^2-a^2b^2c^2 \end{align}

We can do the same thing on the numerator, and the final answer is 1

Gang men
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