If 3 sin (alpha-beta)=5 cos (alpha+beta) and | vedclass

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Question

(C) Given $3 \sin (\alpha-\beta) = 5 \cos (\alpha+\beta)$.Using the expansion formulas,we have $3(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 5

Vedclass · Accepted Answer

$-\frac{1}{4}$

Vedclass · Answer

(C) Given $3 \sin (\alpha-\beta) = 5 \cos (\alpha+\beta)$.Using the expansion formulas,we have $3(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 5(\cos \alpha \cos \beta - \sin \alpha \sin \beta)$.Dividing both sides by $\cos \alpha \cos \beta$,we get $3(	an \alpha - 	an \beta) = 5(1 - 	an \alpha 	an \beta)$.We need to find $X = \frac{	an(\pi/4 - \alpha)}{	an(\pi/4 - \beta)} = \frac{(1 - 	an \alpha)/(1 + 	an \alpha)}{(1 - 	an \beta)/(1 + 	an \beta)} = \frac{(1 - 	an \alpha)(1 + 	an \beta)}{(1 + 	an \alpha)(1 - 	an \beta)}$.From the given equation,$3 	an \alpha - 3 	an \beta = 5 - 5 	an \alpha 	an \beta$,which implies $3 	an \alpha + 5 	an \alpha 	an \beta = 5 + 3 	an \beta$,so $	an \alpha(3 + 5 	an \beta) = 5 + 3 	an \beta$.Thus,$	an \alpha = \frac{5 + 3 	an \beta}{3 + 5 	an \beta}$.Substituting this into the expression for $X$:$X = \frac{(1 - \frac{5 + 3 	an \beta}{3 + 5 	an \beta})(1 + 	an \beta)}{(1 + \frac{5 + 3 	an \beta}{3 + 5 	an \beta})(1 - 	an \beta)} = \frac{(3 + 5 	an \beta - 5 - 3 	an \beta)(1 + 	an \beta)}{(3 + 5 	an \beta + 5 + 3 	an \beta)(1 - 	an \beta)} = \frac{(2 	an \beta - 2)(1 + 	an \beta)}{(8 + 8 	an \beta)(1 - 	an \beta)} = \frac{-2(1 - 	an \beta)(1 + 	an \beta)}{8(1 + 	an \beta)(1 - 	an \beta)} = -\frac{2}{8} = -\frac{1}{4}$.

If $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$ and $\alpha+\beta \neq \frac{\pi}{2}$,then $\frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=$

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