If f(x) = frac{x}{sin x} and g(x) = frac{x}{t | vedclass

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(C) For $f(x) = \frac{x}{\sin x}$,we have $f'(x) = \frac{\sin x - x \cos x}{\sin^2 x} = \frac{\cos x(	an x - x)}{\sin^2 x}$.Since $0 < x \le 1$ (in r

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$f(x)$ is an increasing function

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(C) For $f(x) = \frac{x}{\sin x}$,we have $f'(x) = \frac{\sin x - x \cos x}{\sin^2 x} = \frac{\cos x(	an x - x)}{\sin^2 x}$.Since $0  x$ and $\cos x > 0$. Thus,$f'(x) > 0$,which implies $f(x)$ is an increasing function.For $g(x) = \frac{x}{	an x} = x \cot x$,we have $g'(x) = \cot x - x \csc^2 x = \frac{\sin x \cos x - x}{\sin^2 x} = \frac{\sin 2x - 2x}{2 \sin^2 x}$.Let $h(x) = \sin 2x - 2x$. Then $h'(x) = 2 \cos 2x - 2 = 2(\cos 2x - 1)$. Since $\cos 2x Since $h(0) = 0$ and $h(x)$ is decreasing,$h(x)  0$. Therefore,$g'(x) Thus,$f(x)$ is an increasing function.

If $f(x) = \frac{x}{\sin x}$ and $g(x) = \frac{x}{\tan x}$,where $0 < x \le 1$,then in this interval:

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