If $\alpha, \beta$ are acute angles such that $\sin \beta=2 \sin \alpha$ and $3 \cos \beta=2 \cos \alpha$, then $\sec (\alpha+\beta)=$

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Aicharya · Accepted Answer

Given: $\sin \beta=2 \sin \alpha...(i)$ and $3 \cos \beta=2 \cos \alpha \Rightarrow \cos \beta=\frac{2}{3} \cos \alpha...(ii)$ Now, $\sin ^2 \alpha+\cos ^2 \beta=(2 \sin \alpha)^2+\left(\frac{2}{3} \cos \alphaight)^2$ $\Rightarrow 1=4\left(\sin ^2 \alphaight)+\frac{4}{9}\left(1-\sin ^2 \alphaight)$ $\Rightarrow 1=4 	imes \frac{8}{9} \sin ^2 \alpha+\frac{4}{9} \Rightarrow \sin ^2 \alpha=\frac{5}{32}$ Now, $\sec (\alpha+\beta)=\frac{1}{\cos (\alpha+\beta)}=\frac{1}{\cos \alpha \cos \beta-\sin \alpha \sin \beta}$ $=\frac{1}{\frac{2}{3} \cos ^2 \alpha-2 \sin ^2 \alpha}$ \{from equations (i) \& (ii)\} $=\frac{1}{\frac{2}{3}-\frac{8}{3} \sin ^2 \alpha}=\frac{1}{\frac{2}{3}-\frac{8}{3} \cdot \frac{5}{32}} \Rightarrow \sec (\alpha+\beta)=4$

If $\alpha, \beta$ are acute angles such that $\sin \beta=2 \sin \alpha$ and $3 \cos \beta=2 \cos \alpha$,…

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