If sin x + sin y = frac{sqrt{3}+1}{2} and cos | vedclass

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(A) Given: $\sin x + \sin y = \frac{\sqrt{3}+1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3}-1}{2}$.Using sum-to-product formulas:$2 \sin \left(\frac{x+y

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$8+4 \sqrt{3}$

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(A) Given: $\sin x + \sin y = \frac{\sqrt{3}+1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3}-1}{2}$.Using sum-to-product formulas:$2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{\sqrt{3}+1}{2} \implies \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{\sqrt{3}+1}{4} \quad (i)$$2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{\sqrt{3}-1}{2} \implies \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{\sqrt{3}-1}{4} \quad (ii)$Dividing $(i)$ by $(ii)$:$	an \left(\frac{x+y}{2}ight) = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.So,$	an^2 \left(\frac{x+y}{2}ight) = (2+\sqrt{3})^2 = 4+3+4\sqrt{3} = 7+4\sqrt{3}$.Squaring and adding $(i)$ and $(ii)$:$\cos^2 \left(\frac{x-y}{2}ight) \left[ \sin^2 \left(\frac{x+y}{2}ight) + \cos^2 \left(\frac{x+y}{2}ight) ight] = \left(\frac{\sqrt{3}+1}{4}ight)^2 + \left(\frac{\sqrt{3}-1}{4}ight)^2$$\cos^2 \left(\frac{x-y}{2}ight) = \frac{3+1+2\sqrt{3} + 3+1-2\sqrt{3}}{16} = \frac{8}{16} = \frac{1}{2}$.$\sec^2 \left(\frac{x-y}{2}ight) = 2 \implies 	an^2 \left(\frac{x-y}{2}ight) = 2-1 = 1$.Therefore,$	an^2 \left(\frac{x-y}{2}ight) + 	an^2 \left(\frac{x+y}{2}ight) = 1 + 7 + 4\sqrt{3} = 8+4\sqrt{3}$.

If $\sin x + \sin y = \frac{\sqrt{3}+1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3}-1}{2}$,then $\tan^2 \left(\frac{x-y}{2}\right) + \tan^2 \left(\frac{x+y}{2}\right) = $

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