In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $S=\dfrac13$

Answer 4

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.