The minimum value of $\alpha $, where $\alpha = \frac{4}{sin x}+\frac{1}{1-sinx}$, is

9

The correct answer is option (3) : 9

Let $f(x) = \frac{4}{sin x}+\frac{1}{1-sinx}$. Clearly, f(x) is defined $∀ x ≠ n \pi $ and $ x≠ (4n -1) \frac{\pi }{2} , n \in Z$.

Now,

$f'(x) = - 4cosec x cot x + \frac{cos x}{(1-sinx)^2}=\frac{-4cosx}{sin^2x}+\frac{cos x }{(1-sinx(^2}$

For $ f(x)$ to be maximum or minimum, we must have $f'(x) = 0 $

$⇒\frac{-4cosx}{sin^2x}+\frac{cosx}{(1-sinx)^2}=0$

$⇒\frac{4}{sin^2x}+\frac{cosx}{(1-sinx)^2} $    $[ ∵ cos x ≠0 $ for $x ≠(4n-1)\frac{\pi }{2}]$

$⇒3sin^2x-8sinx + 4= 0 $

$⇒(3sinx - 2) (sinx - 2) = 0 ⇒ 3 sin x -2 =0 ⇒ sinx =\frac{2}{3}$

Now, $f''(x) = 4 cosec x cot^2 x + 4 cosec^3x -\frac{sin x}{(1-sinx)^2}+\frac{2cos^2x}{(1-sinx)^3}$

$⇒f''(\frac{2}{3}) > 0$

Thus, $f(x)$ is minimum when $sin x =\frac{2}{3}$

$∴\alpha = 4×\frac{3}{2}+\frac{1}{(1-2/3)}= 6 + 3= 9 $