A) \[\frac{df}{d\theta }+2f(\theta )cot\theta =0\]
B) \[\frac{df}{d\theta }-2f(\theta )cot\theta =0\]
C) \[\frac{df}{d\theta }+2f(\theta )=0\]
D) \[\frac{df}{d\theta }-2f(\theta )=0\]
Correct Answer: A
Solution :
[a] we have \[f(\theta )=\frac{d}{d\theta }\int\limits_{0}^{\theta }{\frac{dx}{1-\cos \theta \cos x}}\] \[=\frac{1}{1-{{\cos }^{2}}\theta }=\cos e{{c}^{2}}\theta \] (using Leibnitz?s Rule) \[\Rightarrow \frac{df(\theta )}{d\theta }=-2\cos e{{c}^{2}}\theta \cot \theta \] \[\Rightarrow \frac{df(\theta )}{d\theta }+2f(\theta )cot\theta =0\]
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