If cos (A - B) = frac{3}{5} and tan A tan B = | vedclass

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(A) Given $\cos (A - B) = \frac{3}{5}$.Using the expansion formula,$\cos A \cos B + \sin A \sin B = \frac{3}{5}$ ..... $(i)$.Given $	an A 	an B = 2$

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

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(A) Given $\cos (A - B) = \frac{3}{5}$.Using the expansion formula,$\cos A \cos B + \sin A \sin B = \frac{3}{5}$ ..... $(i)$.Given $	an A 	an B = 2$,which implies $\frac{\sin A \sin B}{\cos A \cos B} = 2$,so $\sin A \sin B = 2 \cos A \cos B$ ..... $(ii)$.Substitute $(ii)$ into $(i)$:$\cos A \cos B + 2 \cos A \cos B = \frac{3}{5}$$3 \cos A \cos B = \frac{3}{5}$$\cos A \cos B = \frac{1}{5}$.Now,substitute $\cos A \cos B = \frac{1}{5}$ into $(ii)$:$\sin A \sin B = 2 	imes \frac{1}{5} = \frac{2}{5}$.Thus,the correct option is $\cos A \cos B = \frac{1}{5}$.

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

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