If $f(x) = \frac{x}{\sin x}$ and $g(x) = \frac{x}{\tan x}$, where $0 < x \leq 1$, then in this interval (a) Both $f(x)$ and $g(x)$ are increasing functions (b) Both $f(x)$ and $g(x)$ are decreasing functions (c) $f(x)$ is an increasing function (d) $g(x)$ is an increasing function

1. Setting up the test for monotonicity

For a smooth function, the sign of its derivative decides whether it is increasing or decreasing:

• If $f'(x) \gt 0$ on an interval ➜ $f(x)$ is increasing there.
• If $f'(x) \lt 0$ on an interval ➜ $f(x)$ is decreasing there.

We therefore differentiate both functions.

2. Differentiating $f(x)=\dfrac{x}{\sin x}$

Using the quotient rule $f'(x)=\frac{(\sin x)\cdot 1- x\cdot \cos x}{(\sin x)^2}=\frac{\sin x- x\cos x}{\sin^2 x}$

• Denominator: $(\sin x)^2 \gt 0$ for $0\lt x\le 1$.

• Numerator: Define $h(x)=\sin x- x\cos x$.

• $h(0)=0$.
• $h'(x)=\cos x- \cos x+ x\sin x= x\sin x \ge 0$.
• Hence $h(x)$ is non-decreasing and becomes strictly positive as soon as $x\gt 0$.

Therefore $f'(x)\gt 0$ for $0\lt x\le 1$ ⟹ $f(x)$ is increasing.

3. Differentiating $g(x)=\dfrac{x}{\tan x}$

Again, quotient rule: $g'(x)=\frac{(\tan x)\cdot 1- x\cdot \sec^2 x}{(\tan x)^2}=\frac{\tan x- x\sec^2 x}{\tan^2 x}$

Define $k(x)=\tan x- x\sec^2 x$.

• $k(0)=0$.
• Differentiate:
  $k'(x)=\sec^2 x- \sec^2 x- 2x\sec^2 x\tan x= -2x\sec^2 x\tan x<0\quad (0\lt x\le1)$ So $k(x)$ strictly decreases from 0, making $k(x)<0$ for every $x>0$ in the interval.

Denominator $\tan^2 x>0$, so $g'(x)<0$ ➜ $g(x)$ is decreasing.

4. Comparing with the options

Only the statement (c) "$f(x)$ is an increasing function" is correct.

Imagine two special fraction–machines

1. Machine f(x) takes a number x, puts it on top, and on the bottom it puts the curved-line height of a circle, sin x.
2. Machine g(x) also puts x on top, but on the bottom it puts the slanted-line height, tan x.

We only feed them very small numbers, from a tiny bit above 0 up to 1 (about 57°).

To know whether the machine’s output keeps going up or going down as we turn the dial from 0 to 1, we look at a speed-meter called the derivative:

• A positive speed means the output keeps rising ➜ increasing.
• A negative speed means the output keeps falling ➜ decreasing.

After checking the speed-meters (the derivatives), we discover:

• The speed for f(x) is always positive ➜ it keeps climbing.
• The speed for g(x) is always negative ➜ it keeps sinking.

So only the statement “f(x) is increasing” is true.

Step-by-step derivation

1. For $f(x)=\dfrac{x}{\sin x}$

   f'(x) &= \frac{(\sin x)(1) - x(\cos x)}{(\sin x)^2} \\ &= \frac{\sin x - x\cos x}{\sin^2 x} \end{aligned}$$ Denominator $\sin^2 x > 0$ on $0 < x \le 1$. Let $h(x)=\sin x - x\cos x$. $$h'(x)=x\sin x \ge 0$$ and $h(0)=0$. Therefore $h(x) > 0$ for every $x>0$ in the interval ⟹ $f'(x) > 0$ ⟹ **$f(x)$ increases.**

2. For $g(x)=\dfrac{x}{\tan x}$

   g'(x) &= \frac{(\tan x)(1) - x\sec^2 x}{(\tan x)^2} \\ &= \frac{\tan x - x\sec^2 x}{\tan^2 x} \end{aligned}$$ Denominator $\tan^2 x > 0$ on $0 < x \le 1$. Let $k(x)=\tan x - x\sec^2 x$. $$k'(x)= -2x\sec^2 x\tan x < 0$$ with $k(0)=0$. Therefore $k(x) < 0$ when $x>0$ ⟹ $g'(x) < 0$ ⟹ **$g(x)$ decreases.**

3. Conclusion

   Only option (c) $f(x)$ is an increasing function is correct.