Explanation

(b)
Given, $\cos{x + \sin{x = \frac{1}{2}}}$
$\Rightarrow \ \frac{1 - \tan^{2}\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}} + \frac{2\tan\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}} = \frac{1}{2}$
Let $\tan{\frac{x}{2} = t}$, then
$\frac{1 - t^{2}}{1 + t^{2}} + \frac{2t}{1 + t^{2}} = \frac{1}{2}$
$\Rightarrow \ \ 3t^{2} - 4t - 1 = 0\ \ \ \Rightarrow \ \ t = \frac{2 \pm \sqrt{7}}{3}$
$\Rightarrow \ \ t = \tan{\frac{x}{2} = \frac{2 + \sqrt{7}}{3}}\ \ \ \ \ \ \ \left\lbrack \because 0 < \frac{x}{2} < \frac{\pi}{2},\tan\frac{x}{2}\ is\ positive \right\rbrack$
Now, $\tan{x = \frac{2\tan\frac{x}{2}}{1 - \tan^{2}\frac{x}{2}}}$
$= \frac{2\left( \frac{2 + \sqrt{7}}{3} \right)}{1 - \left( \frac{2 + \sqrt{7}}{3} \right)^{2}} = - \frac{3(2 + \sqrt{7})}{1 + 2\sqrt{7}} \times \frac{1 - 2\sqrt{7}}{1 - 2\sqrt{7}}$
$\Rightarrow \ \tan{x = - \left( \frac{4 + \sqrt{7}}{3} \right)}$