$\lim_{x \to \frac{\pi}{2}}\sin x^{\tan x} = \lim_{x \to \frac{\pi}{2}} e^{\tan x.\ln(\sin x)} \\ e^{\lim_{x \to \frac{\pi}{2}}\tan x.\ln(\sin x) }
\\=e^{\lim_{x \to \frac{\pi}{2}} \frac{\sin x . \ln(\sin x)}{\cos x} \left(\frac{0}{0} \text{ form, applying L'Hospital's rule}\right) }
\\=e^{\lim_{x \to \frac{\pi}{2}} \frac{ \sin x \frac{\cos x}{\sin x} + \ln (\sin x). \cos x}{- \sin x} }
\\= e^0 = 1$