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If $\cos (A - B) = 3 / 5$ and $\tan A \tan B = 2$ , then

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$\cos A \cos B = 1 / 5$

$\sin A \sin B = - 2 / 5$

$\cos A \cos B = - 1 / 5$

$\sin A \sin B = - 1 / 5$

answer is A.

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Detailed Solution

$\cos (A - B) = \frac{3}{5} or 5 \cos A \cos B + 5 \sin A \sin B = 3 . . . . . . (i) .$

From the second relation,

we have $\sin A \sin B = 2 \cos A \cos B . . . . (i i)$

$\Rightarrow \cos A \cos B = \frac{1}{5} and \sin A \sin B = \frac{2}{5}$

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