text{If } sin(alpha+beta)=1, sin(alpha-beta)= | vedclass

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(A) Given $\sin(\alpha+\beta)=1$. Since $\alpha, \beta \in [0, \frac{\pi}{2}]$,$\alpha+\beta = \frac{\pi}{2}$.Given $\sin(\alpha-\beta)=\frac{1}{2}$.

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$1$

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(A) Given $\sin(\alpha+\beta)=1$. Since $\alpha, \beta \in [0, \frac{\pi}{2}]$,$\alpha+\beta = \frac{\pi}{2}$.Given $\sin(\alpha-\beta)=\frac{1}{2}$. Since $\alpha, \beta \in [0, \frac{\pi}{2}]$,$\alpha-\beta = \frac{\pi}{6}$.Adding the two equations: $2\alpha = \frac{\pi}{2} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \implies \alpha = \frac{\pi}{3}$.Subtracting the two equations: $2\beta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \implies \beta = \frac{\pi}{6}$.Now,calculate $	an(\alpha+2\beta) \cdot 	an(2\alpha+\beta)$:$\alpha+2\beta = \frac{\pi}{3} + 2(\frac{\pi}{6}) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$.$2\alpha+\beta = 2(\frac{\pi}{3}) + \frac{\pi}{6} = \frac{2\pi}{3} + \frac{\pi}{6} = \frac{5\pi}{6}$.$	an(\frac{2\pi}{3}) = -\sqrt{3}$ and $	an(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$.Product $= (-\sqrt{3}) \cdot (-\frac{1}{\sqrt{3}}) = 1$.

$\text{If } \sin(\alpha+\beta)=1, \sin(\alpha-\beta)=\frac{1}{2}, \alpha, \beta \in [0, \frac{\pi}{2}], \text{ then } \tan(\alpha+2\beta) \cdot \tan(2\alpha+\beta) = ?$

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