Explanation

(c)
Let $I = \int_{\pi/6}^{\pi/3}\frac{dx}{1 + \sqrt{\cot x}}$
$\Rightarrow \ I = \int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$ …(i)
$\Rightarrow \ I = \int_{\pi/6}^{\pi/3}{\frac{\sqrt{\sin\left( \frac{\pi}{2} - x \right)}}{\sqrt{\sin\left( \frac{\pi}{2} - x \right)} + \sqrt{\cos\left( \frac{\pi}{2} - x \right)}}dx}\ \ \left\lbrack \because\ \int_{a}^{b}{f(x)dx =}\int_{a}^{b}{f(a + b - x)dx} \right\rbrack$
$\Rightarrow \ I = \int_{\pi/6}^{\pi/3}{\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}dx}$ …(ii)
On adding Eqs.(i) and (ii), we get
$2I = \int_{\pi/6}^{\pi/3}{1\ dx = \lbrack x\rbrack_{\pi/6}^{\pi/3}}$
$\Rightarrow I = \frac{1}{2}\left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\pi}{12}$