Explanation

(b)
$\because\ \sec{2\theta = \frac{1}{\cos{2\theta}} = \frac{1 + \tan^{2}\theta}{1 - \tan^{2}\theta}}$
$\therefore\ \tan^{2}{\theta + \sec{2\theta = 1}}$ (given)
$\Rightarrow \ \tan^{2}{\theta + \frac{1 + \tan^{2}\theta}{1 - \tan^{2}\theta} = 1}$
$\Rightarrow \ \tan^{2}{\theta(1 - {tan^{2}}{\theta) + 1 + \tan^{2}{\theta = 1 - \tan^{2}\theta}}}$
$\Rightarrow \ 3\tan^{3}{\theta - \tan^{4}{\theta = 0}}$
$\Rightarrow \ \tan^{2}{\theta(3 - \tan^{2}{\theta) = 0}}$
$\Rightarrow \ \tan{\theta = 0\ or\ \tan{\theta = \pm \sqrt{3}}}$
Now, $\tan{\theta = 0\ \Rightarrow \ \theta =}m\pi$
where $m$ is an integer
And $\tan{\theta = (} \pm \sqrt{3}) = \tan\left( \ \pm \frac{\pi}{3} \right)$
$\Rightarrow \ \theta = n\pi \pm \frac{\pi}{3}$
where $n$ is an integer
Thus, $\theta = m\pi,\ n\pi \pm \frac{\pi}{3}$