A) \[\frac{df}{d\theta }+2f(\theta )\cot \theta =0\]
B) \[\frac{df}{d\theta }-2f(\theta )\cot \theta =0\]
C) \[\frac{2}{3}{{\cos }^{-1}}({{\cos }^{3/2}}x)+c\]
D) None of these
Correct Answer: A
Solution :
\[f(\theta )=\frac{d}{d\theta }\int_{0}^{\theta }{\frac{dx}{1-\cos \theta \cos x}}\] Using Leibnitz's rule, \[f(\theta )=\frac{1}{1-{{\cos }^{2}}\theta }(1-0)\] \[=\frac{1}{1-{{\cos }^{2}}\theta }=\cos e{{c}^{2}}\theta \] On differentiating w.r.t. \[\theta \], we get \[\frac{df}{d\theta }=-2\cos e{{c}^{2}}\theta \cot \theta \] \[\Rightarrow \]\[\frac{df}{d\theta }=2f(\theta )\cot \theta =0\]
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