An extreme value of f(x)=frac{4}{sin x}+frac{ | vedclass

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(A) Given $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.To find the extreme value,we differentiate $f(x)$ with respect to $x$:$f'(x) = -\frac{4 \cos

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$9$

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(A) Given $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.To find the extreme value,we differentiate $f(x)$ with respect to $x$:$f'(x) = -\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2}$.Setting $f'(x) = 0$ for critical points:$-\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2} = 0$.Since $x \in (0, \pi/2)$,$\cos x 
eq 0$,so we can divide by $\cos x$:$-\frac{4}{\sin^2 x} + \frac{1}{(1-\sin x)^2} = 0$.$\frac{1}{(1-\sin x)^2} = \frac{4}{\sin^2 x}$.Taking the square root on both sides:$\frac{1}{1-\sin x} = \frac{2}{\sin x}$ (since $\sin x > 0$ and $1-\sin x > 0$ in the given interval).$\sin x = 2 - 2 \sin x$.$3 \sin x = 2 \Rightarrow \sin x = \frac{2}{3}$.Now,substitute $\sin x = \frac{2}{3}$ into $f(x)$:$f(x) = \frac{4}{2/3} + \frac{1}{1-2/3} = 4 	imes \frac{3}{2} + \frac{1}{1/3} = 6 + 3 = 9$.

An extreme value of $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ in $(0, \pi / 2)$ is

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