(b) Using $\sec 2\theta = \frac{1}{{\cos 2\theta }} = \frac{{1 + {{\tan }^2}\theta }}{{1 – {{\tan }^2}\theta }}$,

we can write the given equation as ${\tan ^2}\theta + \frac{{1 + {{\tan }^2}\theta }}{{1 – {{\tan }^2}\theta }} = 1$.

$ \Rightarrow $ ${\tan ^2}\theta (1 – {\tan ^2}\theta ) + 1 + {\tan ^2}\theta = 1 – {\tan ^2}\theta $

$ \Rightarrow $ $3{\tan ^2}\theta – {\tan ^4}\theta = 0 \Rightarrow {\tan ^2}\theta (3 – {\tan ^2}\theta ) = 0$

$ \Rightarrow $ $\tan \theta = 0$ or $\tan \theta = \pm \sqrt 3 $

Now $\tan \theta = 0 \Rightarrow \theta = m\pi $, where $m$ is an integer and

$\tan \theta = \pm \sqrt 3 = \tan ( \pm \pi /3) \Rightarrow \theta = n\pi \pm \frac{\pi }{3}$, where $n$ is an integer.

Thus $\theta = m\pi ,\,n\pi \pm \frac{\pi }{3}$, where $m$ and $n$ are integers.