Question

In a $△ABC$, if $\cos A\cos B\cos C=\frac{\sqrt{3}-1}{8}$ and $\sin A\sin B\sin C=\frac{3+\sqrt{3}}{8}$, then

The value of $\tan A\tan B+\tan B\tan C+\tan C\tan A$ is

• A. $5-4\sqrt{3}$
• B. $5+4\sqrt{3}$
• C. $6+\sqrt{3}$
• D. $6-\sqrt{3}$

Answer 1

The correct option is B $5+4\sqrt{3}$

$\cos (A+B+C)=\cos A\cos B\cos C[1-\tan A\tan B-\tan B\tan C-\tan C\tan A]\Rightarrow -1=\frac{\sqrt{3}-1}{8}\times [1-\tan A\tan B-\tan B\tan C-\tan C\tan A]\Rightarrow 1-\tan A\tan B-\tan B\tan C-\tan C\tan A=\frac{8}{1-\sqrt{3}}\Rightarrow \tan A\tan B+\tan B\tan C+\tan C\tan A=1-\frac{8}{1-\sqrt{3}}=\frac{-7-\sqrt{3}}{1-\sqrt{3}}=\frac{7+\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=5+4\sqrt{3}$