$\Rightarrow 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{3}{2}$ ..... $(i)$
and $\sin \alpha+\sin \beta=\frac{1}{2}$
$\Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{1}{2}$ ..... $(ii)$
On dividing $(ii)$ by $( i)$, we get
$\tan \left(\frac{\alpha+\beta}{2}\right)=\frac{1}{3}$
Given $: \theta=\frac{\alpha+\beta}{2} \Rightarrow 2 \theta=\alpha+\beta$
Consider $\sin 2 \theta+\cos 2 \theta=\sin (\alpha+\beta)+\cos$
$(\alpha+\beta)$
$=\frac{\frac{2}{3}}{1+\frac{1}{9}}+\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{6}{10}+\frac{8}{10}=\frac{7}{5}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.