If $\sin 2\theta +\sin 2\varphi =\frac{1}{2}$ and $\cos 2\theta +\cos 2\varphi =\frac{3}{2}$ then $co{s}^2(\theta -\varphi )=$

$\sin 2\theta +\sin 2\varphi =\frac{1}{2},\cos 2\theta +\cos 2\varphi =\frac{3}{2}$

$2\sin (\theta +\varphi )\cos (\theta -\varphi )=\frac{1}{2}\cdots \left(1\right)$

$2\cos (\theta +\varphi )\cos (\theta -\varphi )=\frac{3}{2}\cdots \left(2\right)$

$4\sin (\theta +\varphi )\cos (\theta +\varphi ){\cos }^2(\theta -\varphi )=\frac{1}{2}\times \frac{3}{2}$

$2(2\sin (\theta +\varphi )\cos (\theta +\varphi \left)\right){\cos }^2(\theta -\varphi )=\frac{3}{4}$

$2\sin 2(\theta +\varphi ){\cos }^2(\theta -\varphi )=\frac{3}{4}$

$2\left(\frac{2\tan (\theta +\varphi )}{1+{\tan }^2(\theta +\varphi )}\right){\cos }^2\left(\theta -\varphi \right)=\frac{3}{4}$

$\frac{4\times {\displaystyle \frac{1}{3}}}{1+{\left({\displaystyle \frac{1}{3}}\right)}^2}{\cos }^2\left(\theta -\varphi \right)=\frac{3}{4}$

$\frac{{\displaystyle \frac{4}{3}}}{1+{\displaystyle \frac{1}{9}}}{\cos }^2\left(\theta -\varphi \right)=\frac{3}{4}$

$\frac{4}{3}.\frac{9}{10}{\cos }^2\left(\theta -\varphi \right)=\frac{3}{4}$

$\frac{6}{5}{\cos }^2(\theta -\varphi )=\frac{3}{4}$

$co{s}^2(\theta -\varphi )=\frac{3}{4}\cdot \frac{5}{6}$

$=\frac{5}{8}$