MCQ
If $\cos (A - B) = \frac{3}{5}$ and $\tan A\tan B = 2,$ then
- ✓$\cos A\cos B = \frac{1}{5}$
- B$\sin A\sin B = - \frac{2}{5}$
- C$\cos A\cos B = - \frac{1}{5}$
- D$\sin A\sin B = - \frac{1}{5}$
✓
Answer
Correct option: A.
$\cos A\cos B = \frac{1}{5}$
a
(a) $\cos \,(A - B) = \frac{3}{5}$
(a) $\cos \,(A - B) = \frac{3}{5}$
$\therefore$ $5\,\,\cos A\,\,\cos B + 5\,\,\sin A\,\,\sin B = 3$…..$(i) $
From $2^{nd}$ relation, $\sin A\sin B = 2\cos A\cos B$ .....$(ii)$
$\therefore $ $\cos A\cos B = \frac{1}{5}$
and $5\,\left( {\frac{1}{2} + 1} \right)\,\sin A\,\,\sin B = 3$.
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