Explanation

(c)
Since, $a\cos{x + b\sin{x = c}}$
$\therefore\ a\frac{1 - \tan^{2}\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}} + b\frac{2\tan\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}} = c$
$\Rightarrow \ a - {a\ tan}^{2}{\frac{x}{2} + 2b\tan{\frac{x}{2} = c}}\left( 1 + \tan^{2}\frac{x}{2} \right)$
$\Rightarrow \ (c + a)\tan^{2}{\frac{x}{2} - 2b\tan{\frac{x}{2} + c - a = 0}}$
Since, $\alpha,\ \beta$ are both roots of the given equation
$\therefore\ \tan{\frac{\alpha}{2} + \tan{\frac{\beta}{2} = \frac{2b}{c + a}}}$
and $\tan{\frac{\alpha}{2} + \tan{\frac{\beta}{2} = \frac{c - a}{c + a}}}$
Now, $\tan{\left( \frac{\alpha + \beta}{2} \right) = \frac{\tan{\frac{\alpha}{2} + \tan\frac{\beta}{2}}}{1 - \tan{\frac{\alpha}{2}\ \bullet \ \tan\frac{\beta}{2}}}\ }$
$\Rightarrow \ \tan{\left( \frac{\alpha + \beta}{2} \right) = \frac{\frac{2b}{c + a}}{1 - \frac{c - a}{c + a}}}$
$\Rightarrow \ \tan{\left( \frac{\alpha + \beta}{2} \right) = \frac{b}{a}}$