Question:
If $\cos (A-B)=\frac{3}{5}$ and $\tan A \tan B=2$, then $\sin A \sin B=$ ______________ .
Solution:
$\cos (A-B)=\frac{3}{5}$ Given
and $\tan A \tan B=2$
Since $\cos (A-B)=3 / 5$
i. e. $\cos A \cos B+\sin A \sin B=3 / 5$
i. e. $\cos A \cos B\left(1+\frac{\sin A \sin B}{\cos A \cos B}\right)=3 / 5$
i. e. $\cos A \cos B(1+\tan A \tan B)=3 / 5$
i.e. $\cos A \cos B(1+2)=3 / 5$
$\cos A \cos B=\frac{3}{5} \times \frac{1}{3}$
$\cos A \cos B=\frac{1}{5}$
$\therefore \sin A \sin B=\frac{3}{5}-\frac{1}{5}$
$\sin A \sin B=\frac{2}{5}$
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