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If $f(x)=\frac{x}{\sin x},g(x)=\frac{x}{\tan x}, 0< x \leq 1$, then in this interval (1) both $f(x), g(x)$ are increasing functions (2) both $f(x), g(x)$ are decreasing functions (3) $f(x)$ is increasing function (4) $g(x)$ is increasing function.

Trial: $f'(x)=\frac{\sin x - x\cos x}{\sin^2x}$ and $g'(x)=\frac{\tan x -x\sec^2x}{\tan^2x}=\frac{\sin x\cos x-x}{\sin^2x}$

From here I can't conclude anything.please help.

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up vote 2 down vote accepted

Hint: Remember you are working in $(0,1)$.

Let $h(x)=\sin x-x\cos x$. Then $h(0)=0$ and $$h^{\prime}(x)=\cos x-\cos x+x\sin x=x\sin x> 0$$ in $(0,1)$. Can you continue?

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Can I do this since $\sin^2x > 0$ in the given interval? – Argha Dec 30 '12 at 12:23
@Argha Because $(0,1)\subset (0,\pi)$ $\sin x> 0$ in $(0,1)$ – Nameless Dec 30 '12 at 12:25
This answer is clear to me. Is there another way to solve this problem? – Argha Dec 30 '12 at 12:28
@Argha You mean without calculus? – Nameless Dec 30 '12 at 12:29
yes you are right – Argha Dec 30 '12 at 12:31

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