Question

In a $\Delta ABC$, if cosA cosB cosC = $\frac{\sqrt{3}-1}{8}$ and sinA sinB sinC = $\frac{3+\sqrt{3}}{8}$ , then
The value of tan A tan B + tan B tan C + tan C tan A is

Answer 1

$=\tan A\tan B+\tan B\tan C+\tan C\tan A=\frac{\sin A\sin B\sin C+\sin B\sin C\cos A+\sin C\sin A\cos B}{\cos A\cos B\cos C}=3(\sin A\sin B\sin C+\sin C(\sin B\cos A+\sin A\cos B\left)\right)=3(\sin A\sin B\sin C+\sin C(\sin (A+B)\left)\right)=3(\sin A\sin B\sin C+\sin {}^{2}A)=3(\sin A\sin B\sin C+1-\cos {}^{2}C)=3(1+\cos C(-\cos C+\sin A\sin B\left)\right)=3(1+\cos C\cos B\cos A)=3(1+\frac{1}{3})=4$