A geometry problem by A Former Brilliant Member

Using half-angle tangent substitution , we have $\sin x = \dfrac {2t}{1+t^2}$ and $\cos x = \dfrac {1-t^2}{1+t^2}$ , where $t = \tan \frac x2$ . Then:

$\begin{aligned} \sqrt{\frac {1+\sin x}{1-\sin x}} + \sqrt{\frac {1-\sin x}{1+\sin x}} & = \sqrt{\frac {1+\frac {2t}{1+t^2}}{1-\frac {2t}{1+t^2}}} + \sqrt{\frac {1-\frac {2t}{1+t^2}}{1+\frac {2t}{1+t^2}}} \\ & = \sqrt{\frac {t^2 + 2t+1}{t^2 - 2t+1}} + \sqrt{\frac {t^2 - 2t+1}{t^2 + 2t+1}} \\ & = \frac {t+1}{t-1} + \frac {t-1}{t+1} \\ & = \frac {2(t^2+1)}{t^2-1} = - \frac {2(1+t^2}{1-t^2} = \boxed{- \dfrac 2{\cos x}} \end{aligned}$

$\begin{aligned} &\frac{1+\sin(x)}{1-\sin(x)} {\color{#D61F06} \cdot \frac{1+\sin(x)}{1+\sin(x)}}=\frac{1+2 \sin(x)+\sin^2(x)}{\cos^2(x)} =\sec^2(x)+ 2\sec(x) \tan(x) + \tan^2(x)=(\sec(x)+tan(x))^2 \\ &\frac{1-\sin(x)}{1+\sin(x)} {\color{#D61F06} \cdot \frac{1-\sin(x)}{1-\sin(x)}}=\frac{1-2 \sin(x)+\sin^2(x)}{\cos^2(x)} =\sec^2(x)- 2\sec(x) \tan(x) + \tan^2(x)=(\sec(x)-tan(x))^2 \\ & Now\\ &\sqrt{\frac{1+\sin(x)}{1-\sin(x)}} + \sqrt{\frac{1-\sin(x)}{1+\sin(x)}}=\sqrt{(\sec(x)+tan(x))^2} + \sqrt{(\sec(x)-tan(x))^2} = |\sec(x)+tan(x)| + |\sec(x)-tan(x)|\\ & \color{#D61F06} \frac{\pi}{2}< x < \pi \Rightarrow \sec(x)+tan(x) \leq 0 \Rightarrow |\sec(x)+tan(x)| = -(\sec(x)+tan(x)) \\ & \color{#D61F06} \frac{\pi}{2}< x < \pi \Rightarrow \cos(x)<0 , \sin(x) \leq 1 \Rightarrow \frac{\sin(x)}{\cos(x)} \geq \frac{1}{\cos(x)} \Rightarrow \sec(x)-tan(x) \leq 0 \Rightarrow |\sec(x)-tan(x)|= - (\sec(x)-tan(x)) \\ & so \\ &|\sec(x)+tan(x)| + |\sec(x)-tan(x)| = -(\sec(x)+tan(x)) - (\sec(x)-tan(x)) = -2 \sec(x) = - \frac{2}{\cos(x)}\\ \end{aligned}$