The function ⨍ (θ)=ddθ∫0θdx1-cosθ cosx satisfies the differential equation

f(θ)=ddθ∫0θdx1− cosθ cosx=11−cosθ cosθ=11−cosθ=1sin2θf(θ)=cosec2θ⇒ ddθ(f(θ))=−2cosec2θ.cotθ⇒ ddθ(f(θ))+2f(θ)cotθ=0

The function ⨍ (θ)=ddθ∫0θdx1-cosθ cosx satisfies the differential equation

f(θ)=ddθ∫0θdx1− cosθ cosx=11−cosθ cosθ=11−cosθ=1sin2θf(θ)=cosec2θ⇒ ddθ(f(θ))=−2cosec2θ.cotθ⇒ ddθ(f(θ))+2f(θ)cotθ=0

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The function $⨍ (θ)= \frac{d}{dθ} \int_{0}^{θ} \frac{dx}{1 - cosθ cosx}$ satisfies the differential equation

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$\frac{d ⨍}{dθ} - 2 ⨍ (θ)cotθ=0$

$\frac{d ⨍}{dθ} - 2 ⨍ (θ)=0$

$\frac{d ⨍}{dθ} + 2 ⨍ (θ)cotθ=0$

$\frac{d ⨍}{dθ} + 2 ⨍ (θ)=0$

answer is A.

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Detailed Solution

$\begin{array}{l} f (θ) = \frac{d}{d θ} \int_{0}^{θ} \frac{d x}{1 - \cos θ \cos x} \\ = \frac{1}{1 - \cos θ \cos θ} \\ = \frac{1}{1 - \cos θ} \\ = \frac{1}{\sin^{2} θ} \\ f (θ) = \cos e c^{2} θ \\ \Rightarrow \frac{d}{d θ} (f (θ)) = - 2 \cos e c^{2} θ . \cot θ \\ \Rightarrow \frac{d}{d θ} (f (θ)) + 2 f (θ) \cot θ = 0 \end{array}$