The limit generate an uncertain form of the type $[\infty^{0}]$ and that is pretty much I can say about it.
I know that $\cos x$ goes to $\frac{\pi}{2}^{-}$ as $-x+\frac{\pi}{2}$ but I do not see how it helps in moving away from $[\infty^{0}]$.
Any suggestion?
By using the substitution $y=\cos(x)$ you get: $$\lim_{x\to\frac{\pi}{2}^-}\left(\frac{1}{cos(x)}\right)^{\cos(x)}=\lim_{y\to 0^+}\left(\frac{1}{y}\right)^y =\lim_{y\to 0^+}(y)^{-y}.$$ Using the continuity of logarithm we get: $$\ln(\lim_{y\to 0^+}(y)^{-y})= \lim_{y\to 0^+}\ln((y)^{-y})=\lim_{y\to 0^+}-y\ln(y)=\lim_{y\to 0^+}-\frac {\ln y}{\frac{1}{y}}\stackrel{\text{L'Hôpital's rule}}{=}0$$ $\implies \ln(\lim_{y\to 0^+}(y)^{-y})=0\implies \lim_{y\to 0^+}(y)^{-y}=1$
If you want to go beyond the limit, let $x=\frac \pi 2-t$.
$$y=\big(\csc(t) \big)^{\sin(t)}\quad \implies \log(y)=\sin(t)\,\log\big(\csc(t) \big)$$ Now, using Taylor around $t=0^+$ $$\csc(t)=\frac{1}{t}+\frac{t}{6}+\frac{7 t^3}{360}+O\left(t^5\right)$$ $$\log\big(\csc(t) \big)=-\log (t)+\frac{t^2}{6}+\frac{t^4}{180}+O\left(t^6\right)$$ Multiply by the usual series expansion of $\sin(t)$ $$\log(y)=-t \log (t)+\frac{1}{6} t^3 (\log (t)+1)-\frac{1}{360} t^5 (3 \log (t)+8)+O\left(t^7\right)$$ Now $$y=e^{\log(y)}=1-t \log (t)+\frac{1}{2} t^2 \log ^2(t)+\cdots$$
Compare the plots to see how good (or bad) is the approximation
