# Prove that $$f(\theta)$$ = $${4sin \theta\over 2 + cos\theta} – \theta$$ is an increasing function of $$\theta$$ in $$[0, {\pi\over 2}]$$.

## Solution :

We have, $$f(\theta)$$ = $${4sin \theta\over 2 + cos\theta} – \theta$$

$$\implies$$ $$f'(\theta)$$ = $$(2 + cos\theta)(4 cos\theta) + 4 sin^2\theta\over (2 + cos\theta)^2$$ – 1

$$\implies$$ $$f'(\theta)$$  = $$8 cos\theta + 4\over (2 + cos\theta)^2$$ – 1

$$\implies$$ $$f'(\theta)$$ = $$4\cos\theta – cos^2\theta\over (2 + cos\theta)^2$$

$$\implies$$ $$f'(\theta)$$ = $$cos\theta(4 – cos\theta)\over (2 + cos\theta)^2$$ > 0  for all $$\theta$$ $$\in$$ $$(0, {\pi\over 2})$$.

[ $$\because$$   $$cos\theta$$ > 0 , 4 – $$cos\theta$$ > 0 and 2 + $$cos\theta$$ > 0 ]

Hence, $$f(\theta)$$ is increasing on $$[0, {\pi\over 2}]$$.

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