જો (alpha+beta) એ frac{pi}{2} નો ગુણક ન હોય અ | vedclass

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(A) આપેલ છે કે $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$.પદોનું વિસ્તરણ કરતા,$3(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 5(\cos \alpha \

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(A) આપેલ છે કે $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$.પદોનું વિસ્તરણ કરતા,$3(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 5(\cos \alpha \cos \beta - \sin \alpha \sin \beta)$ મળે.પદોને ગોઠવતા,$\sin \alpha(3 \cos \beta + 5 \sin \beta) = \cos \alpha(5 \cos \beta + 3 \sin \beta)$ મળે.તેથી,$	an \alpha = \frac{5 \cos \beta + 3 \sin \beta}{3 \cos \beta + 5 \sin \beta} = \frac{5 + 3 	an \beta}{3 + 5 	an \beta}$.હવે,$	an \left(\frac{\pi}{4} + \alphaight) = \frac{1 + 	an \alpha}{1 - 	an \alpha} = \frac{1 + \frac{5 + 3 	an \beta}{3 + 5 	an \beta}}{1 - \frac{5 + 3 	an \beta}{3 + 5 	an \beta}} = \frac{8(1 + 	an \beta)}{-2(1 - 	an \beta)} = -4 	an \left(\frac{\pi}{4} + \betaight)$.તેથી,$	an \left(\frac{\pi}{4} + \alphaight) + 4 	an \left(\frac{\pi}{4} + \betaight) = 0$.

જો $(\alpha+\beta)$ એ $\frac{\pi}{2}$ નો ગુણક ન હોય અને $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$ હોય,તો $\tan \left(\frac{\pi}{4}+\alpha\right)+4 \tan \left(\frac{\pi}{4}+\beta\right)=$

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