If sin x + sin y = alpha and cos x + cos y = | vedclass

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(C) Given $\sin x + \sin y = \alpha$ and $\cos x + \cos y = \beta$. Using sum-to-product formulas: $2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac

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$\frac{\alpha^2 + \beta^2}{2 \alpha \beta}$

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(C) Given $\sin x + \sin y = \alpha$ and $\cos x + \cos y = \beta$. Using sum-to-product formulas: $2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \alpha$ $2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \beta$ Dividing the two equations: $	an \left(\frac{x+y}{2}ight) = \frac{\alpha}{\beta}$ Using the identity $\sin(x+y) = \frac{2 	an \left(\frac{x+y}{2}ight)}{1 + 	an^2 \left(\frac{x+y}{2}ight)}$: $\sin(x+y) = \frac{2(\alpha/\beta)}{1 + (\alpha/\beta)^2} = \frac{2 \alpha \beta}{\beta^2 + \alpha^2}$ Therefore,$\operatorname{cosec}(x+y) = \frac{1}{\sin(x+y)} = \frac{\alpha^2 + \beta^2}{2 \alpha \beta}$.

If $\sin x + \sin y = \alpha$ and $\cos x + \cos y = \beta$,then $\operatorname{cosec}(x + y) = $

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