Given, $\sin A+\sin B=a$
$$
\cos A+\cos B=b
$$
Dividing Eq. (i) by Eq. (ii), we get
$$
\begin{array}{ll}
& \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{a}{b} \\
\Rightarrow \quad & \frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}=\frac{a}{b} \\
\Rightarrow \quad & \tan \frac{A+B}{2}=\frac{a}{b} \\
\text { Now, } \quad & \cos (A+B)=\frac{1-\tan ^2\left(\frac{A+B}{2}\right)}{1+\tan ^2\left(\frac{A-B}{2}\right)}
\end{array}
$$

$$
\left(\because \text { using } \cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right)
$$

$$
\begin{aligned}
& \Rightarrow \quad \cos (A+B)=\frac{1-\frac{a^2}{b^2}}{1+\frac{a^2}{b^2}} \\
& \therefore \quad \cos (A+B)=\frac{b^2-a^2}{b^2+a^2}
\end{aligned}
$$
[Using Eq. (iii)]