Question

The function $f\left(\theta \right)=\frac{d}{d\theta }{\int }_0^{\theta }\frac{dx}{1-\cos \theta \cos x}$ satisfies the differential equation

• A. $\frac{df}{d\theta }+2f\left(\theta \right)\cot \theta =0$
• B. $\frac{df}{d\theta }-2f\left(\theta \right)\cot \theta =0$
• C. $\frac{df}{d\theta }+2f\left(\theta \right)=0$
• D. $\frac{df}{d\theta }-2f\left(\theta \right)=0$

Answer 1

The correct option is A $\frac{df}{d\theta }+2f\left(\theta \right)\cot \theta =0$

$f\left(\theta \right)=\frac{d}{d\theta }{\int }_0^{\theta }\frac{dx}{1-\cos \theta \cos x}$
$\Rightarrow f\left(\theta \right)=\frac{1}{1-{\cos }^{\theta }}\frac{d}{d\theta }\left(\theta \right)-\frac{1}{1-\cos \theta }\frac{d}{d\theta }\left(0\right)$
$\Rightarrow f\left(\theta \right)=\frac{1}{{\sin }^2\theta }$ .....(1)
Differentiating w.r.t $\theta$, we get
$\Rightarrow \frac{d}{d\theta }f\left(\theta \right)=\frac{-2}{{\sin }^3\theta }\cos \theta$
$\Rightarrow \frac{d}{d\theta }f\left(\theta \right)=-2f\left(\theta \right)\cot \theta$
$\Rightarrow \frac{d}{d\theta }f\left(\theta \right)+2f\left(\theta \right)\cot \theta =0$