Idea for $\lim\limits_{x \to \frac{\pi}{2}} \left( \tan \left( \tfrac{\pi}{4} \sin x\right)\right)^{1/ ( \tan (\pi \sin x))}$ $$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.
ps. can't use lhopital

Using trigonometric identities, by $y=\frac \pi 2 \sin x \to \frac \pi 2$ and $t=\frac \pi 2 -y \to 0$, we obtain:


*

*$\tan \frac{\theta}{2}  = \frac{\sin \theta}{1 +
   \cos \theta} \implies \tan \left(\frac{\pi}{4}\sin x\right)=\frac{\sin y}{1 +
   \cos y}=\frac{\cos t}{1 +
   \sin t}$

*$\tan (2\theta) = \frac{2 \tan \theta} {1 - \tan^2 \theta} \implies \tan(\pi \sin x)=\frac{2 \tan y} {1 - \tan^2 y}=\frac{2 \cot t} {1 - \cot^2 t}=\frac{2\cos t \sin t}{\sin^2 t-\cos^2t}$
then
$$\tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=e^{{\dfrac {\log \left(\tan \left(\frac{\pi}{4}\sin x\right)\right)}{\tan(\pi \sin x)} }}
=e^{\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\log\left(\frac{\cos t}{1 +
   \sin t} \right)}\to e^\frac12$$
indeed
$$\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot \log\left(\frac{\cos t}{1 +
   \sin t} \right)=\\=\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot\frac{\cos t-\sin t-1}{1+\sin t}\cdot\log\left(1+\frac{\cos t-\sin t-1}{1 +
   \sin t} \right)^{\frac{1+\sin t}{\cos t-\sin t -1}}\to\frac 12 \cdot \log e=\frac 12$$
since
$$\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot\frac{\cos t-\sin t-1}{1+\sin t}=\frac{\cos^2t-\sin^2 t}{\cos t(1+\sin t)}\cdot\frac{\sin t+1-\cos t}{2\sin t}=1\cdot \frac12$$
and by standard limits
$$\frac{\sin t+1-\cos t}{2\sin t}=\frac{ \frac{\sin t}{t}+t\frac{1-\cos t}{t^2} } {2\frac {\sin t} t}\to \frac{1+0\cdot 1}{2\cdot 1}=\frac12$$
