Question

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

• A. $2$
• B. $1$
• C. $3$
• D. $-2$

Answer 1

The correct option is A $2$

$\because A+B+C=\pi \Rightarrow C=\pi -(A+B)\Rightarrow -\cos (A+B)=\cos C...\left(1\right)\therefore \frac{\cos C}{\cos A\cos B}=-\frac{\cos (A+B)}{\cos A\cos B}\Rightarrow \frac{\cos C}{\cos A\cos B}=-\frac{\cos A\cos B-\sin A\sin B}{\cos A\cos B}\Rightarrow \frac{\cos C}{\cos A\cos B}=\tan A\tan B-1\Rightarrow \frac{\cos C}{\cos A\cos B}=2$