Explanation

(a)
$a\cos{\theta + b\sin{\theta = c}}$ …(i)
$\because\ \ \alpha$ and $\beta\ (\alpha \neq \beta)$ satisfy the Eq. (i)
$\Rightarrow a\cos{\alpha + b\sin{\alpha = c}}$ …(ii)
And $a\cos{\beta + b\sin{\beta = c}}$ …(iii)
On, subtracting Eq. (iii) from Eq.; (ii), we get
$a\cos{\alpha +}b\sin{\alpha - a\cos{\beta - b\sin{\beta = 0}}}$
$\Rightarrow \ a{(cos}{\alpha - \cos{\beta)} = - b}(\sin{\alpha - \sin\beta})$
$\Rightarrow \ \ a\sin\frac{\alpha + \beta}{2} = - b\left\lbrack - \cos\left( \frac{\alpha + \beta}{2} \right) \right\rbrack$
$\Rightarrow \ \tan\left( \frac{\alpha + \beta}{2} \right) = \frac{b}{a}$