Explanation

(a)
Since, $\sin{A,\sin{B,\cos A}}$ are in GP
$\therefore\ \sin^{2}{B = \sin{A\cos A}}$ ...(i)
Also, $x^{2} + 2x\cot{B + 1 = 0}$ [given]
Now, $b^{2} - 4ac = 4\cot^{2}{B - 4}$
$= \frac{4\cos^{2}{B - 4\sin^{2}B}}{\sin^{2}B}$
$= \frac{4(1 - 2\sin^{2}B)}{\sin^{2}B}$
$= \frac{4(1 - 2\sin{A\cos A})}{\sin^{2}B}$
$= 4\left( \frac{\sin{A - \cos A}}{\sin B} \right)^{2}$ [from Eq. (i)]
$\geq 0$
$\therefore$ Roots of given equation are always real