If cos (A-B)=3/5 and tan A tan B=2,then which | vedclass

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(D) Given,$\cos (A-B)=3/5$ and $	an A 	an B=2$. We know that $	an A 	an B = \frac{\sin A \sin B}{\cos A \cos B} = 2$. Applying componendo and divi

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$\cos (A+B)=-1/5$

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(D) Given,$\cos (A-B)=3/5$ and $	an A 	an B=2$. We know that $	an A 	an B = \frac{\sin A \sin B}{\cos A \cos B} = 2$. Applying componendo and dividendo: $\frac{\cos A \cos B + \sin A \sin B}{\cos A \cos B - \sin A \sin B} = \frac{2+1}{2-1}$. This simplifies to $\frac{\cos (A-B)}{\cos (A+B)} = 3$. Substituting $\cos (A-B) = 3/5$: $\frac{3/5}{\cos (A+B)} = 3$. $\cos (A+B) = \frac{3/5}{3} = 1/5$. Wait,let us re-evaluate the sign: $\frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B + \sin A \sin B} = \frac{1-2}{1+2} = -1/3$. $\frac{\cos (A+B)}{\cos (A-B)} = -1/3$. $\cos (A+B) = -1/3 	imes 3/5 = -1/5$.

If $\cos (A-B)=3/5$ and $\tan A \tan B=2$,then which one of the following is true?

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