(c) $\sec \theta + \tan \theta = \sqrt 3 $….$(i)$

Also we have ${\sec ^2}\theta – {\tan ^2}\theta = 1$…$(ii)$

$ \Rightarrow $ $\sec \theta – \tan \theta = \frac{1}{{\sqrt 3 }}$…$(iii)$

Now $(i)$ and $(iii)$ gives,

$\tan \theta = \frac{1}{2}\left( {\sqrt 3 – \frac{1}{{\sqrt 3 }}} \right) = \frac{1}{{\sqrt 3 }} = \tan \left( {\frac{\pi }{6}} \right)$

$ \Rightarrow $ $\theta = n\pi + \frac{\pi }{6}$.

$\therefore $ Solutions for $0 \le \theta \le 2\pi $ are $\frac{\pi }{6}$ and $\frac{{7\pi }}{6}$.

Hence there are two solutions.