Explanation

(c)
$\because\sec\theta + \tan{\theta = \sqrt{3}}$ …(i)
Also, we have
$\sec^{2}{\theta - \tan^{2}{\theta = 1}}$ …(ii)
$\Rightarrow \ (\sec{\theta + \tan{\theta)(\sec{\theta - \tan{\theta) = 1}}}}$
$\Rightarrow \ \sec{\theta - \tan{\theta = \frac{1}{\sqrt{3}}}}$ …(iii)
From Eqs. (i) and (iii), we get
$\tan{\theta = \frac{1}{2}\left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = \frac{1}{\sqrt{3}} = \tan\frac{\pi}{6}\ }$
$\Rightarrow \ \theta = n\pi + \frac{\pi}{6}$
$\therefore$ Solutions for $0 < \theta < 2\pi\ are\ \frac{\pi}{6}\ and\ \frac{7\pi}{6}$
Hence, there are two solutions