If int_0^{frac{pi}{2}} frac{cot x}{cot x+oper | vedclass

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(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x$. Simplifying the integrand: $\frac{\cot x}{\cot x+\operatorname{

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$-1$

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(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x$. Simplifying the integrand: $\frac{\cot x}{\cot x+\operatorname{cosec} x} = \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} = \frac{\cos x}{\cos x+1}$. Now,$I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{1+\cos x} d x = \int_0^{\frac{\pi}{2}} \left( \frac{1+\cos x - 1}{1+\cos x} ight) d x = \int_0^{\frac{\pi}{2}} \left( 1 - \frac{1}{1+\cos x} ight) d x$. Using the identity $1+\cos x = 2\cos^2 \frac{x}{2}$,we get $I = \int_0^{\frac{\pi}{2}} \left( 1 - \frac{1}{2}\sec^2 \frac{x}{2} ight) d x$. Integrating,we get $I = \left[ x - 	an \frac{x}{2} ight]_0^{\frac{\pi}{2}} = \left( \frac{\pi}{2} - 	an \frac{\pi}{4} ight) - (0 - 	an 0) = \frac{\pi}{2} - 1$. We can write this as $\frac{1}{2}(\pi - 2)$. Comparing with $m(\pi+n)$,we get $m = \frac{1}{2}$ and $n = -2$. Therefore,$m \cdot n = \frac{1}{2} \cdot (-2) = -1$.

If $\int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=m(\pi+n)$,then $(m \cdot n)$ equals

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