Question:
If $\cos (A-B)=\frac{3}{5}$ and $\tan A \tan B=2$, then
(a) $\cos A \cos B=\frac{1}{5}$
(b) $\cos A \cos B=-\frac{1}{5}$
(c) $\sin A \sin B=-\frac{1}{5}$
(d) $\sin A \sin B=-\frac{1}{5}$
Solution:
(a) $\frac{1}{5}$
$\tan A \tan B=\frac{\sin A \sin B}{\cos A \cos B}=2 \quad$ (Given) $\quad \ldots$ (1)
Also,
$\cos (A-B)=\frac{3}{5}$
$\Rightarrow \cos A \cos B+\sin A \sin B=\frac{3}{5}$
$\therefore \sin A \sin B=\frac{3}{5}-\cos A \cos B \quad \ldots(2)$
Substituting eq (2) in eq (1), we get:
$\Rightarrow \frac{\frac{3}{5}-\cos A \cos B}{\cos A \cos B}=2$
$\Rightarrow 3 \cos A \cos B=\frac{3}{5}$
$\Rightarrow \cos A \cos B \quad=\frac{1}{5}$