If the sum of the square of the roots of the equation $x^{2} - (\sin\alpha - 2)x - (1+\sin\alpha) = 0$ is least, then $\alpha$ is equal to (1) $\frac{\pi}{6}$ (2) $\frac{\pi}{4}$ (3) $\frac{\pi}{3}$ (4) $\frac{\pi}{2}$
If the sum of the square of the roots of the equation $x^{2} - (\sin\alpha - 2)x - (1+\sin\alpha) = 0$ is least, then $\alpha$ is equal to\\
(1) $\frac{\pi}{6}$\\
(2) $\frac{\pi}{4}$\\
(3) $\frac{\pi}{3}$\\
(4) $\frac{\pi}{2}$