Q.Prove that $y = \frac{4\sin\theta}{(2+\cos\theta)} - \theta$ is an increasing function of $\theta$ in $[0, \frac{\pi}{2}]$.

To prove a function is increasing on an interval, show $y'\ge 0$ there.

Step 1 — Differentiate (quotient rule on the first term):

The fraction's numerator is $8\cos\theta+4\cos^2\theta+4\sin^2\theta=8\cos\theta+4$.

Step 2 — Combine over one denominator:

$y'=\frac{8\cos\theta+4-(2+\cos\theta)^2}{(2+\cos\theta)^2}=\frac{4\cos\theta-\cos^2\theta}{(2+\cos\theta)^2}=\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}.$ …

The idea

A function is increasing on an interval when its derivative is non-negative throughout. So we compute $y'$, simplify it to a single fraction, and check its sign on $[0,\frac{\pi}{2}]$.

Set up

Work the steps

1. Differentiate the quotient. With $u=4\sin\theta$, $v=2+\cos\theta$, so $u'=4\cos\theta$, $v'=-\sin\theta$:

The numerator is

using $\sin^2\theta+\cos^2\theta=1$. The derivative of $-\theta$ is $-1$, so

2. Combine into one fraction by writing $1=\dfrac{(2+\cos\theta)^2}{(2+\cos\theta)^2}$:

Expand $(2+\cos\theta)^2=4+4\cos\theta+\cos^2\theta$, so the numerator is

Hence

$y'=\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}.$ …

Method: Proving Monotonicity on a Closed Interval Using the Quotient Rule and a Trig Identity

For a function combining a trigonometric fraction with a linear term (like $\theta$), proving monotonicity on a specific closed interval combines two skills: correctly applying the quotient rule, and then using a Pythagorean identity to collapse the messy result into something whose sign is clear on that particular interval.

Steps

Step 1: Apply the quotient rule to the fractional trig term

For a term of the form $\dfrac{a\sin\theta}{b+\cos\theta}$, use $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$, then differentiate the remaining linear term (like $-\theta$) separately and subtract $1$.

Step 2: Expand the numerator and apply $\sin^2\theta + \cos^2\theta = 1$

The numerator from the quotient rule typically contains both a $\cos^2\theta$ and a $\sin^2\theta$ term — recognising and substituting the Pythagorean identity is what collapses the expression into a simple polynomial in $\cos\theta$ alone. Skipping this step leaves an expression whose sign looks impossible to determine.

Step 3: Combine everything into a single fraction over a common denominator

Rewrite the $-1$ using the same denominator as the quotient-rule term, then simplify the combined numerator fully — factor it if possible.

Step 4: Determine the sign of each factor specifically on the given closed interval …

Common Mistakes

Mistake 1: Forgetting to apply $\sin^2\theta + \cos^2\theta = 1$ to simplify the numerator

Why it's wrong: After applying the quotient rule, the numerator contains both a $\cos^2\theta$ and a $\sin^2\theta$ term; without substituting the Pythagorean identity, the expression looks messy and its sign is not obviously determinable, which can make a student wrongly conclude the proof is "stuck." Correct approach: always look for $\sin^2\theta + \cos^2\theta$ appearing after expanding a quotient-rule numerator involving both sine and cosine, and replace it with $1$ immediately.

Mistake 2: Assuming $\cos\theta$ can be negative on $[0, \frac{\pi}{2}]$ …