Explanation

(b)
On squaring given equation, we get
$\sin^{2}{A + 6\cos^{2}{A - 2\sqrt{6}\sin A\cos{A = 7\cos^{2}A}}}$
$\Rightarrow \ \sin^{2}A + 6\left( 1 - \sin^{2}A \right) = \cos^{2}{A + 6\cos^{2}{A + 2\sqrt{6}}\sin A}\cos A$
$\Rightarrow \sin^{2}{A - 6}\cos^{2}{A + 6 = \cos^{2}{A + 6\sin^{2}{A + 2\sqrt{6}\sin A}}}\cos A$
$\Rightarrow \ 7{\sin^{2}{A = \left( \cos{A + \sqrt{6}}\sin A \right)}}^{2}$
$\Rightarrow \ \ \pm \sqrt{7}\sin{A = \cos{A + \sqrt{6}}}\sin A$
Alternate
Given, $\sin A - \sqrt{6}\cos{A = \sqrt{7}}\cos A$
Replacing $A$ by $90{^\circ} + A$, we get
$\sin{(90{^\circ} + A)} - \sqrt{6}\cos{(90{^\circ} + A)}$
$= \sqrt{7}\cos{(90{^\circ} + A)}$
$\Rightarrow \cos{A + \sqrt{6}}\sin A = - \sqrt{7}\sin A$