(a)
We have, $\tan{n\theta = \tan{m\theta}}$
$\Rightarrow n\theta = N\pi + (m\theta)$
$\Rightarrow \ \theta = \frac{N\pi}{n - m},$ putting $N = 1,\ 2,\ 3,\ldots$ we get
$\frac{\pi}{n - m},\frac{2\pi}{n - m},\frac{3\pi}{n - m}\ldots$ which are obviously in AP.