If

$\int_{0}^{\pi / 2} \frac{\cot x d x}{\cot x+\operatorname{cosec} x}$

$=\int_{0}^{\pi / 2} \frac{\cot x d x}{1+\cos x}=\int_{0}^{\pi / 2}\left(1-\frac{1}{1+\cos x}\right) d x$

$=[x]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \frac{1}{2 \cos ^{2} \frac{x}{2}} d x=\frac{\pi}{2}-\frac{1}{2} \int_{0}^{\pi / 2} \sec ^{2} \frac{x}{2} d x$

$=\frac{\pi}{2}-\left(\tan \frac{x}{2}\right)_{0}^{\pi / 2}=\frac{\pi}{2}-[1]=\left(\frac{\pi}{2}-1\right)=m \pi+m n$

$\therefore m=\frac{1}{2}, n=-2$, Hence, $m n=-1$