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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