Solution:

Given, $ a=\sin x+\sin y b=\cos x+\cos y\\ $

$$ \begin{aligned} &\therefore \mathrm{a}^{2}+\mathrm{b}^{2} \\\\ &=[\sin \mathrm{x}+\sin \mathrm{y}]^{2}+[\cos \mathrm{x}+\cos \mathrm{y}]^{2} \\\\ &\text { Formula } \mathrm{S}+\mathrm{S}=2 \mathrm{SC} \quad \mathrm{C}+\mathrm{C}=2 \mathrm{CC} \\\\ &\left.=\left[2 \sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\right]^{2}+2 \cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\right]^{2} \\\\ &=4 \sin ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)+4 \cos ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right) \\\\ &4 \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right) \text { is } \operatorname{common} \text { to both } \operatorname{are} \operatorname{terms} \\\\ &=4 \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\left[\sin ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)+\cos ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\right] \\\\ &=4 \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right) \times 1\quad \because \sin ^{2} \theta+\cos ^{2} \theta=1 \\\\ &\mathrm{a}^{2}+\mathrm{b}^{2}=4 \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right) \\\\ \end{aligned} $$

$$ \begin{aligned} \text { RHS } &=\frac{4-a^{2}-b^{2}}{\left(a^{2}+b^{2}\right)}=\frac{4-\left(a^{2}+b^{2}\right)}{\left(a^{2}+b^{2}\right)} \\\\ &=\frac{4-4 \cos ^{2}\left(\frac{x-y}{2}\right)}{4 \cos ^{2}\left(\frac{x-y}{2}\right)}=\frac{4\left[1-\cos ^{2}\left(\frac{x-y}{2}\right)\right]}{4 \cos ^{2}\left(\frac{x-y}{2}\right)} \\\\ &=\frac{\sin ^{2}\left(\frac{x-y}{2}\right)}{\cos ^{2}\left(\frac{x-y}{2}\right)} \quad \because 1-\cos ^{2} \theta=\sin ^{2} \theta \\\\ &=\tan ^{2}\left(\frac{x-y}{2}\right)=\text { RHS } \end{aligned} $$