lf x and y are acute angles, such that $\cos \mathrm{x}+\cos \mathrm{y}=\frac{3}{2}$ and $\sin \mathrm{x}+\sin \mathrm{y}=\frac{3}{4}$ then sin (x + y) equals

$\begin{array}{l}\text{ Given, }\cos \mathrm{x}+\cos \mathrm{y}=\frac{3}{2}\\ \Rightarrow 2\cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\frac{3}{2}----\left(\mathrm{i}\right)\\ \text{ and }\sin \mathrm{x}+\sin \mathrm{y}=\frac{3}{4}\\ \Rightarrow 2\sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\frac{3}{4}----\left(\mathrm{ii}\right)\end{array}$

On dividing Eq. (ii) by Eq. (i), we get

$\begin{array}{l}\frac{2\sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)}{2\cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)}=\frac{3/4}{3/2}\Rightarrow \tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=\frac{1}{2}\\ \therefore \sin (\mathrm{x}+\mathrm{y})=\frac{2\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)}{1+{\tan }^2\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)}=\frac{2\times \frac{1}{2}}{1+{\left(\frac{1}{2}\right)}^2}=\frac{4}{4+1}=\frac{4}{5}\end{array}$