$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x} \biggl| $ 

$$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{\cos^2x} \biggl| $$


My Approach:
$$\lim_{x \to \frac{\pi}{2}} \biggl|\frac{1}{(1+\sin x)\log(\sin x)}+\frac{1}{(1-\sin x)(1+\sin x)} \biggl|$$
$$\lim_{x \to \frac{\pi}{2}} \frac{1}{2}\biggl|\frac{1}{\log(\sin x)}+\frac{1}{(1-\sin x)} \biggl| $$
$$\lim_{x \to \frac{\pi}{2}} \frac{1}{2}\biggl|\frac{(\sin x - 1)}{(\sin x-1)\log(1+(\sin x - 1))}+\frac{1}{(1-\sin x)} \biggl|$$
I could not processed further.
By Applying  L'Hôpital's Rule I get answer $0$ but given answer is $\frac{1}{4}$
Also L'Hôpital's Rule is going lengthy.

Let me use $\ln$ in place of $\log$ and get from your third line
$$\frac{1}{\ln(\sin x)}+\frac{1}{(1-\sin x)} =\frac{1-\sin x+\ln(\sin x)}{(1-\sin x)\ln(\sin x)} = \\=
\frac{1-\sin x+\sin x -1+(\sin x-1)^2\frac{1}{2}-(\sin x - 1)^3\frac{1}{3}+\cdots}{-(\sin x - 1)^2}\to -\frac{1}{2}$$
