To prove that $y = \dfrac{4 \sin \theta}{2 + \cos \theta} - \theta$ is an increasing function of $\theta$ in $\left[0, \dfrac{\pi}{2}\right]$, we need to show that $\dfrac{dy}{d\theta} > 0$ in this interval.

Step 1: Find the derivative

$$y = \dfrac{4 \sin \theta}{2 + \cos \theta} - \theta$$

Using quotient rule for the first term:

$$\frac{dy}{d\theta} = \frac{(2 + \cos \theta)(4 \cos \theta) - (4 \sin \theta)(-\sin \theta)}{(2 + \cos \theta)^2} - 1$$ $$= \frac{4(2 + \cos \theta)\cos \theta + 4 \sin^2 \theta}{(2 + \cos \theta)^2} - 1$$ $$= \frac{8 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta}{(2 + \cos \theta)^2} - 1$$ $$= \frac{8 \cos \theta + 4(\cos^2 \theta + \sin^2 \theta)}{(2 + \cos \theta)^2} - 1$$ $$= \frac{8 \cos \theta + 4}{(2 + \cos \theta)^2} - 1$$

Step 2: Simplify the derivative

$$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4 - (2 + \cos \theta)^2}{(2 + \cos \theta)^2}$$ $$= \frac{8 \cos \theta + 4 - 4 - 4 \cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2}$$ $$= \frac{4 \cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2}$$ $$= \frac{\cos \theta(4 - \cos \theta)}{(2 + \cos \theta)^2}$$

Step 3: Analyze the sign in $\left[0, \dfrac{\pi}{2}\right]$

For $\theta \in \left[0, \dfrac{\pi}{2}\right]$:

• $\cos \theta \geq 0$ (cosine is non-negative in first quadrant)
• $4 - \cos \theta > 0$ (since $\cos \theta \leq 1 < 4$)
• $(2 + \cos \theta)^2 > 0$ (always positive)

Therefore, $\dfrac{dy}{d\theta} > 0$ for all $\theta \in \left[0, \dfrac{\pi}{2}\right]$.

Conclusion: Since the derivative is positive throughout the interval $\left[0, \dfrac{\pi}{2}\right]$, the function $y$ is an increasing function of $\theta$ in $\left[0, \dfrac{\pi}{2}\right]$.