Question

In a $\Delta ABC$ , if $\cos A\cos B\cos C=\frac{1}{3}$, then the value of $\tan A\tan B+\tan B\tan C+\tan C\tan A$ is

• A. $1$
• B. $\frac{4}{3}$
• C. $4$
• D. $3$

Answer 1

The correct option is C $4$

To find $\tan A\tan B+\tan B\tan C+\tan C\tan A$

$=\frac{\sin A\sin B}{\cos A\cos B}+\frac{\sin B\sin C}{\cos B\cos C}+\frac{\sin C\sin A}{\cos C\cos A}$

$=\frac{\sin A\sin B\cos C+\cos A\sin B\sin C+\sin C\sin A\cos B}{\cos A\cos B\cos C}$

$=\frac{-\cos A\cos B\cos C+\cos (A+B+C)}{\cos A\cos B\cos C}$

As $\cos (A+B+C)=\sin A\sin B\cos C+\sin A\cos B\sin C+\cos A\sin B\sin C+\cos A\cos B\cos C$
And $A+B+C=180\Rightarrow \cos (A+B+C)=-1$

$\therefore \frac{-\cos A\cos B\cos C+\cos (A+B+C)}{\cos A\cos B\cos C}=\frac{-\frac{1}{3}-1}{-\frac{1}{3}}=4$