If $(\alpha+\beta)$ is not a multiple of $\frac{\pi}{2}$ and $3 \sin (\alpha-\beta)=5$ $\cos (\alpha+\beta)$, then $ an \left(\frac{\pi}{4}+\alpha ight)+4 an \left(\frac{\pi}{4}+\beta ight)=$

Question

If $(\alpha+\beta)$ is not a multiple of $\frac{\pi}{2}$ and $3 \sin (\alpha-\beta)=5$ $\cos (\alpha+\beta)$, then $	an \left(\frac{\pi}{4}+\alphaight)+4 	an \left(\frac{\pi}{4}+\betaight)=$

Aicharya · Accepted Answer

$3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$ $\begin{aligned} & 3 \sin \alpha \cos \beta-3 \cos \alpha \sin \beta=5 \cos \alpha \cos \beta-5 \sin \alpha \sin \beta \ & \sin \alpha(3 \cos \beta+5 \sin \beta)=\cos \alpha(5 \cos \beta+3 \sin \beta) \ & 	an \alpha=\frac{5 \cos \beta+3 \sin \beta}{3 \cos \beta+5 \sin \beta}=\frac{5+3 	an \beta}{3+5 	an \beta} \ & 	an \left(\frac{\pi}{4}+\alphaight)=\frac{1+	an \alpha}{1-	an \alpha}=\frac{\left[1+\frac{5+3 	an \beta}{3+5 	an \beta}ight]}{1-\frac{5+3 	an \beta}{3+5 	an \beta}} \ & =\frac{8(1+	an \beta)}{-2(1-	an \beta)}=-4 	an \left(\frac{\pi}{4}+\betaight) \ & 	an \left(\frac{\pi}{4}+\alphaight)+4 	an \left(\frac{\pi}{4}+\betaight)=0\end{aligned}$

If $(\alpha+\beta)$ is not a multiple of $\frac{\pi}{2}$ and $3 \sin (\alpha-\beta)=5$ $\cos (\alpha+\beta)$…

Solution