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 $\tan A+\tan B+\tan C$ is

• A. $\frac{3+\sqrt{3}}{\sqrt{3}-1}$
• B. $\frac{\sqrt{3}+4}{\sqrt{3}-1}$
• C. $\frac{6-\sqrt{3}}{\sqrt{3}-1}$
• D. $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-1}$

Answer 1

The correct option is A $\frac{3+\sqrt{3}}{\sqrt{3}-1}$

$\because A,B,C$ are angles of triangle,
$\therefore \tan A+\tan B+\tan C=\tan A\tan B\tan C=\frac{\sin A\sin B\sin C}{\cos A\cos B\cos C}=\frac{3+\sqrt{3}}{\sqrt{3}-1}$