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In $\triangle ABC$, if $$\cos A \cos B \cos C=\frac{1}{3}$$ then can we find value of $$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$

Please give some hint. I am not sure if $\tan A \tan B+\tan B \tan C+\tan C \tan A$ will be constant under given condition.

user26857
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Ananya
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  • Possibly use cos(A+B) = cos(A).cos(B) - sin(A).sin(B), and similar formula for cos(π-C) remembering, for a triangle A+B+C = π and then simplyfying. – Arif Burhan Apr 02 '16 at 17:15
  • The accepted answer is incorrect, because there is no triangle $\triangle ABC$ with $\cos A \cos B \cos C=\frac 13$. See my answer to this question. – Steve Kass Apr 05 '16 at 19:04

4 Answers4

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Let $$S=\tan A\tan B+\tan B\tan C+\tan C\tan A$$

Multiplying by $\cos A \cos B \cos C=\frac 13$, we get $$\frac 13S=\sin A\sin B\cos C+\cos A\sin B\sin C+\sin A\cos B \sin C$$

However,$$\cos(A+B+C)=-1=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C$$

Therefore, $$\frac 13S=\cos A\cos B\cos C+1=\frac 43\Rightarrow S=4$$

David Quinn
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  • Writing the formula for S as a function of cos(A)cos(B)cos(C), which follows from the same argument, would be more illuminating. – zyx Apr 02 '16 at 17:58
  • @zyx can you show us what you mean? – David Quinn Apr 02 '16 at 18:25
  • I meant to replace 1/3 by a variable $u$ and write S in terms of $u$. Your method does this. If you prefer that I edit your post to include that, I can do so. – zyx Apr 02 '16 at 19:25
  • There is nothing wrong with your algebra, but I think there is no triangle for which $\cos A \cos B \cos C=\frac{1}{3}$. See my answer. – Steve Kass Apr 05 '16 at 17:35
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The answer to the question is no. The maximum value of $\cos A \cos B \cos C$, where $A$, $B$, and $C$ are the angles of a triangle in the plane, is $\frac{1}{8}$, so there is no plane triangle for which $\cos A \cos B \cos C=\frac{1}{3}$.

The product $\cos A \cos B \cos C$ equals $\frac{1}{8}$ for an equilateral triangle, and the fact that this is a maximum follows from the fact that for any acute* triangle $\triangle ABC$, the product is greater for the “more equilateral” triangle with angles $\frac{A+B}{2}$, $\frac{A+B}{2}$, and $C$, because $$\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)\cos C-\cos A\cos B\cos C= \frac{1-\cos(A+B)}{2}\cdot\cos C>0.$$

*We can assume the triangle is acute (and $\cos C>0$), because otherwise the product $\cos A \cos B \cos C \le 0$ (only one angle in a given triangle can be non-acute, so only one of the cosines can be non-positive) and $\triangle ABC$ can’t possibly be one for which $\cos A \cos B \cos C$ is a maximum.

This might be an interesting question for triangles on a surface of negative curvature, however.

Steve Kass
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Hint: For a triangle ABC

$A+B=\pi-C$

and

$1-2\cos A \cos B \cos C=\cos^2A+\cos^2B+\cos^2C$

Edit $1$: $A+B=\pi-C$

Apply $\cos$ on both sides

Divide each term by $\cos A.\cos B$

We get $\tan A.\tan B=1+\frac{\cos C}{\cos A. \cos B}$

Similarly write $2$ more equations and add three equations.

Now use $1-2\cos A \cos B \cos C=\cos^2A+\cos^2B+\cos^2C$

Mathematics
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Rephrasing Mathematics's answer:

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ and $\cos(A+B)=\cdots=-\cos C$

$\cos^2A+\cos^2B+\cos^2C=1+\cos^2A-\sin^2B+\cos^2C=1-2\cos A\cos B\cos C$

If $\cos A\cos B\cos C=S$

$\iff\cos^2A+\cos^2B+\cos^2C=1-2S$ let $y=\tan A\tan B\iff y-1=-\dfrac{\cos(A+B)}{\cos A\cos B}=\dfrac{\cos^2C}S$

$\iff\cos^2C=S(y-1)$

$\sum S(y-1)=\sum\cos^2C=1-2S\iff y=3+\dfrac{1-2S}S$

Here $S=\dfrac13$

Community
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lab bhattacharjee
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  • Unfortunately, there is no triangle $\triangle ABC$ for which $\cos A\cos B\cos C=\frac{1}{3}$. – Steve Kass Apr 05 '16 at 19:11
  • @SteveKass, Observe that my answer is not dependent on the actual value of $\cos A\cos B\cos C$ – lab bhattacharjee Apr 06 '16 at 16:30
  • @SteveKass, Simpler proof : If $y=\cos A\cos B\cos C,$ $$\iff2y=2\cos A\cos B\cos C=\cos A[\cos(B-C)+\cos(B+C)]$$

    As $B+C=\pi-A,$ $$2y=\cos A[\cos(B-C)-\cos A]$$

    $$\iff\cos^2A-\cos A\cos(B-C)+2y=0$$

    As $\cos A$ is real, the discriminant $\cos^2(B-C)-8y\ge0\iff8y\le\cos^2(B-C)$

    The equality occurs if $\cos^2(B-C)=1=\cos0\iff B=C$ as $0<B,C<\pi$

    In that case, $$\cos^2A-\cos A+2\cdot\dfrac18=0$$ $$\iff\cos A=\dfrac12\implies A=60^\circ\implies B+C=20^\circ$$ But $B=C\implies B=C=60^\circ=A$

     – lab bhattacharjee Apr 06 '16 at 16:36