If alpha, beta are acute angles such that sin | vedclass

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(A) Given: $\sin \beta=2 \sin \alpha \dots (i)$ and $3 \cos \beta=2 \cos \alpha \Rightarrow \cos \beta=\frac{2}{3} \cos \alpha \dots (ii)$ Squaring an

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$4$

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(A) Given: $\sin \beta=2 \sin \alpha \dots (i)$ and $3 \cos \beta=2 \cos \alpha \Rightarrow \cos \beta=\frac{2}{3} \cos \alpha \dots (ii)$ Squaring and adding the equations: $\sin^2 \beta + \cos^2 \beta = (2 \sin \alpha)^2 + (\frac{2}{3} \cos \alpha)^2$ $1 = 4 \sin^2 \alpha + \frac{4}{9} \cos^2 \alpha$ $1 = 4 \sin^2 \alpha + \frac{4}{9} (1 - \sin^2 \alpha)$ $1 = 4 \sin^2 \alpha + \frac{4}{9} - \frac{4}{9} \sin^2 \alpha$ $1 - \frac{4}{9} = \frac{32}{9} \sin^2 \alpha$ $\Rightarrow \frac{5}{9} = \frac{32}{9} \sin^2 \alpha$ $\Rightarrow \sin^2 \alpha = \frac{5}{32}$ Similarly,$\cos^2 \alpha = 1 - \frac{5}{32} = \frac{27}{32}$ Now,$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ Substitute $\cos \beta = \frac{2}{3} \cos \alpha$ and $\sin \beta = 2 \sin \alpha$: $\cos(\alpha + \beta) = \cos \alpha (\frac{2}{3} \cos \alpha) - \sin \alpha (2 \sin \alpha)$ $\cos(\alpha + \beta) = \frac{2}{3} \cos^2 \alpha - 2 \sin^2 \alpha$ $\cos(\alpha + \beta) = \frac{2}{3} (\frac{27}{32}) - 2 (\frac{5}{32}) = \frac{18}{32} - \frac{10}{32} = \frac{8}{32} = \frac{1}{4}$ Therefore,$\sec(\alpha + \beta) = \frac{1}{\cos(\alpha + \beta)} = 4$.

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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