If x in (0, frac{pi}{4}),then the expression | vedclass

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(A) Let $f(x) = \frac{\cos x}{\sin^2 x(\cos x - \sin x)}$.We can rewrite the denominator as $\sin x \cdot \sin x(\cos x - \sin x)$.Using the $AM \geq

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$8$

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(A) Let $f(x) = \frac{\cos x}{\sin^2 x(\cos x - \sin x)}$.We can rewrite the denominator as $\sin x \cdot \sin x(\cos x - \sin x)$.Using the $AM \geq GM$ inequality for the terms $\sin x$ and $(\cos x - \sin x)$:$\frac{\sin x + (\cos x - \sin x)}{2} \geq \sqrt{\sin x(\cos x - \sin x)}$$\Rightarrow \frac{\cos x}{2} \geq \sqrt{\sin x(\cos x - \sin x)}$Squaring both sides:$\frac{\cos^2 x}{4} \geq \sin x(\cos x - \sin x)$Now,substitute this into the expression:$f(x) = \frac{\cos x}{\sin x \cdot \sin x(\cos x - \sin x)} \geq \frac{\cos x}{\sin x \cdot \frac{\cos^2 x}{4}} = \frac{4}{\sin x \cos x} = \frac{8}{2 \sin x \cos x} = \frac{8}{\sin 2x}$.Since $x \in (0, \frac{\pi}{4})$,$2x \in (0, \frac{\pi}{2})$,so $\sin 2x \in (0, 1)$.Therefore,$\frac{8}{\sin 2x} > 8$.This implies that the expression $f(x)$ must be strictly greater than $8$. Thus,it cannot take the value $8$ or any value less than $8$.

If $x \in (0, \frac{\pi}{4})$,then the expression $\frac{\cos x}{\sin^2 x(\cos x - \sin x)}$ cannot take the value

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