If cos (A − B) =

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:

$\Rightarrow \frac{\frac{3}{5}-\cos A \cos B}{\cos A \cos B}=2$

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

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

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