If m cos (alpha+beta)-n cos (alpha-beta)=m co | vedclass

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(C) Given: $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$Rearranging the terms,we get:$m [\cos (\alpha+\bet

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$-\frac{n}{m}$

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(C) Given: $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$Rearranging the terms,we get:$m [\cos (\alpha+\beta) - \cos (\alpha-\beta)] = n [\cos (\alpha+\beta) + \cos (\alpha-\beta)]$Using the identities $\cos (A+B) - \cos (A-B) = -2 \sin A \sin B$ and $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$:$m [-2 \sin \alpha \sin \beta] = n [2 \cos \alpha \cos \beta]$$-2m \sin \alpha \sin \beta = 2n \cos \alpha \cos \beta$Dividing both sides by $2m \cos \alpha \cos \beta$ (assuming $\cos \alpha \cos \beta 
eq 0$):$\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = -\frac{n}{m}$$	an \alpha 	an \beta = -\frac{n}{m}$

If $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$,then $\tan \alpha \tan \beta=$

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