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) $\sin A\sin B=-\frac{1}{5}$

Answer 1

(a) $\frac{1}{5}$
$\tan A\tan B=\frac{\sin A\sin B}{\cos A\cos B}=2\left(\mathrm{Given}\right)...\left(1\right)$

$$\begin{gathered} \text{Also,} \\ \cos (A-B)=\frac{3}{5} \\ \Rightarrow \cos A\cos B+\sin A\sin B=\frac{3}{5} \end{gathered}$$
$$\begin{gathered} \therefore \sin A\sin B=\frac{3}{5}-\cos A\cos B...\left(2\right) \\ \mathrm{Substituting}\mathrm{eq}\left(2\right)\mathrm{in}\mathrm{eq}\left(1\right),\mathrm{we}\mathrm{get}: \end{gathered}$$
$\Rightarrow \frac{{\displaystyle \frac{3}{5}}-\cos A\cos B}{\cos A\cos B}=2$

$\Rightarrow 3\cos A\cos B={\displaystyle \frac{3}{5}}$

$\Rightarrow \cos A\cos B=\frac{1}{5}$