$\begin{aligned} & I=\int_0^\pi \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} d x \\ & \qquad \because \text { using } \\ & \qquad \int_0^a f(x) d x=\int_0^a f(a-x) d x\end{aligned}$
$=\int_0^\pi \frac{-(\pi-x) \tan (x)}{-\sec x-\tan x} d x$

Adding Eqs. (i) and (ii)
$$
2 I=\int_0^\pi \frac{\pi \tan x}{\sec x+\tan x} d x=\pi \int_0^\pi \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} d x
$$
$\begin{aligned} &=\pi \int_0^\pi \frac{\sin x}{1+\sin x} d x=\pi \int_0^\pi \frac{\sin x+1-1}{1+\sin x} d x \\ &=\pi\left[\int_0^\pi \frac{(1+\sin x)}{1+\sin x} d x-\int_0^\pi \frac{1}{1+\sin x} d x\right] \\ &=\pi\left[\int_0^\pi 1 d x-\int_0^\pi \frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x\right] \\ &=\pi\left[[x]_0^\pi-\int_0^\pi \frac{1-\sin x}{\cos ^2 x} d x\right]=\pi[(\pi-0)- \\ & \int_0^\pi\left(\sec ^2 x-\sec x \tan x\right) d x=\pi\left[\pi-(\tan x-\sec x)_0^\pi\right] \\ &=\pi[\pi-[\tan \pi-\sec \pi-\tan 0+\sec 0]] \\ &=\pi[\pi-\{0+1-0+1\}] \\ & \quad 2 I=\pi[\pi-2] \Rightarrow I=\frac{\pi}{2}[\pi-2]\end{aligned}$