If int frac{1}{1-cos x} dx = tan left(frac{x} | vedclass

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(B) Given the integral: $\int \frac{1}{1-\cos x} dx = \int \frac{1}{2 \sin^2 \frac{x}{2}} dx$ $= \frac{1}{2} \int \operatorname{cosec}^2 \frac{x}{2} d

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$\pi$

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(B) Given the integral: $\int \frac{1}{1-\cos x} dx = \int \frac{1}{2 \sin^2 \frac{x}{2}} dx$ $= \frac{1}{2} \int \operatorname{cosec}^2 \frac{x}{2} dx$ $= \frac{1}{2} \left(-2 \cot \frac{x}{2}ight) + c = -\cot \frac{x}{2} + c$ We know that $-\cot 	heta = 	an \left(	heta + \frac{\pi}{2}ight)$ or $	an \left(	heta - \frac{\pi}{2}ight)$. Using $-\cot \frac{x}{2} = 	an \left(\frac{x}{2} - \frac{\pi}{2}ight)$,we compare this with $	an \left(\frac{x}{\alpha} + \betaight)$. Thus,$\alpha = 2$ and $\beta = -\frac{\pi}{2}$. Calculating the required value: $\frac{\pi \alpha}{4} - \beta = \frac{\pi(2)}{4} - \left(-\frac{\pi}{2}ight) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.

If $\int \frac{1}{1-\cos x} dx = \tan \left(\frac{x}{\alpha} + \beta\right) + c$,then one of the values of $\frac{\pi \alpha}{4} - \beta$ is

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