Explanation

(a)
Since, $\tan{\theta + \tan{\left( \frac{3\pi}{4} + \theta \right) = 2}}$
$\therefore\ \tan{\theta + \frac{- 1 + \tan\theta}{1 + \tan\theta} = 2}$
$\Rightarrow \ \tan{\theta + \tan^{2}{\theta - 1 + \tan{\theta = 2 + 2\tan\theta}}}$
$\Rightarrow \ \tan^{2}\theta = 3$
$\Rightarrow \ \tan^{2}{\theta = \left( \ \sqrt{3} \right)^{2} = \tan^{2}\frac{\pi}{3}}$
$\Rightarrow \ \theta = n\pi \pm \frac{\pi}{3},\ n \in I$