If $\sin\alpha+\sin\beta=a,\cos\alpha+\cos\beta=b$ then $\cos(\alpha+\beta)$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{b^2-a^2}{b^2+a^2}&(b)\;\large\frac{b^2+a^2}{b^2+a^2}\\(c)\;\large\frac{a^2-b^2}{b^2+a^2}&(d)\;None\;of\;these\end{array}$

We have
$b^2+a^2=(\cos\alpha+\cos \beta)^2+(\sin\alpha+\sin\beta)^2$
$\qquad\quad=(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)$
$\qquad\quad=1+1+2\cos(\alpha-\beta)$
$\qquad\quad=2+2\cos(\alpha-\beta)$-------(1)
Step 2:
$b^2-a^2=(\cos\alpha+\cos\beta)^2-(\sin\alpha+\sin\beta)^2$
$\qquad\quad=\cos^2\alpha+\cos^2\beta-\sin^2\alpha-\sin^2\beta+2(\cos\alpha\cos\beta-\sin\alpha\sin\beta)$
$\qquad\quad=(\cos^2\alpha-\sin^2\beta)+(\cos^\beta-\sin^2\alpha)+2\cos(\alpha+\beta)$
$\qquad\quad=\cos(\alpha+\beta)\cos(\alpha-\beta)+\cos(\beta+\alpha)\cos(\beta-\alpha)+2\cos(\alpha+\beta)$
$\qquad\quad=2\cos(\alpha+\beta)\cos(\alpha-\beta)+2\cos(\alpha+\beta)$
$\qquad\quad=\cos(\alpha+\beta)[2\cos(\alpha-\beta)+2]$
By using equ (1)
$\qquad\quad=\cos(\alpha+\beta)(b^2+a^2)$
Step 3:
Thus we have
$b^2-a^2=(b^2+a^2)\cos(\alpha+\beta)$
$\Rightarrow \cos(\alpha+\beta)=\large\frac{b^2-a^2}{b^2+a^2}$
Hence (a) is the correct answer.