Break the limit

$\begin{aligned} L & = \lim_{x \to \frac \pi 2} \frac {4(x-\pi)\cos^2 x}{\pi(\pi-2x)\color{#3D99F6} \tan \left(x - \frac \pi 2\right)} \\ & = \lim_{x \to \frac \pi 2} \frac {4(x-\pi)\cos^2 x}{\pi(\pi-2x)\color{#3D99F6}\left(- \tan \left(\frac \pi 2 - x \right)\right)} \\ & = \lim_{x \to \frac \pi 2} {\color{#3D99F6} -} \frac {4(x-\pi)\cos^2 x}{\pi(\pi-2x)\color{#3D99F6}\cot x} \\ & = \lim_{x \to \frac \pi 2} - \frac {{\color{#3D99F6}4}(x-\pi)\color{#3D99F6}\sin x \cos x}{\pi(\pi-2x)} \\ & = \lim_{x \to \frac \pi 2} - \frac {{\color{#3D99F6}2}(x-\pi)\color{#3D99F6}\sin 2x}{\pi(\pi-2x)} & \small \color{#3D99F6} \text{Note that }\sin (\pi - x) = \sin x \\ & = \lim_{x \to \frac \pi 2} - \frac {2(x-\pi)\sin (\pi - 2x)}{\pi(\pi-2x)} & \small \color{#3D99F6} \text{Let } u = \pi - 2x \\ & = \lim_{u \to 0} \frac {(\pi + u)\color{#3D99F6}\sin u}{\pi \color{#3D99F6}u} & \small \color{#3D99F6} \text{Note that } \lim_{x \to 0} \frac {\sin x}x = 1 \\ & = \boxed{1} \end{aligned}$

$\lim_{x \rightarrow \frac{\pi}{2}} \frac{4 (x-\pi) \cos^{2} x}{\pi (\pi-2x) \tan (x-\frac{\pi}{2})}$ $=\lim_{x \rightarrow \frac{\pi}{2}} \frac{2 (x-\pi) \sin^{2} (\frac{\pi}{2}-x)}{\pi (\frac{\pi}{2}-x) \tan (x-\frac{\pi}{2})}$ $=\lim_{x \rightarrow \frac{\pi}{2}} \frac{2 (\pi-x) \sin (\frac{\pi}{2}-x) \cdot \sin (\frac{\pi}{2}-x)}{\pi (\frac{\pi}{2}-x) \tan (\frac{\pi}{2}-x)}$ $=\lim_{x \rightarrow \frac{\pi}{2}} \frac{2 (\pi-x)}{\pi}$ $=\frac{2 (\pi-\frac{\pi}{2})}{\pi}=\frac{2 (\frac{\pi}{2})}{\pi}=\boxed{1}$