(c) $a\cos x + b\sin x = c$

$ \Rightarrow $ $a\,\left( {\frac{{1 – {{\tan }^2}\,(x/2)}}{{1 + {{\tan }^2}(x/2)}}} \right) + \frac{{2b\tan (x/2)}}{{1 + {{\tan }^2}(x/2)}} = c$

$ \Rightarrow $ $(a + c){\tan ^2}\frac{x}{2} – 2b\tan \frac{x}{2} + (c – a) = 0$

This equation has roots $\tan \frac{\alpha }{2}$ and $\tan \frac{\beta }{2}$.

$\therefore $ $\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} = \frac{{2b}}{{a + c}}$ and

$\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{{c – a}}{{a + c}}$

Now $\tan \left( {\frac{\alpha }{2} + \frac{\beta }{2}} \right) = \frac{{\tan \frac{\alpha }{2} + \tan \frac{\beta }{2}}}{{1 – \tan \frac{\alpha }{2}\tan \frac{\beta }{2}}}$

$= \frac{{\frac{{2b}}{{a + c}}}}{{1 – \frac{{c – a}}{{a + c}}}} = \frac{b}{a}$.