Calculating $\lim_{x\to\frac{\pi}{2}}(\sin x)^{\tan x}$ Please help me calculate: $\lim_{x\to\frac{\pi}{2}}(\sin x)^{\tan x}$

Take $y=(\sin x)^{\tan x}$
Taking log on both sides we have,
$\log y=\tan x\log(\sin x)=\frac{\log(\sin x)}{\cot x}$
Now as $x\to \pi/2$, $\log(\sin x)\to 0$ and $\cot x\to 0$
Now you can use L'Hospital's Rule.
$$\lim_{x\to \pi/2}\frac{\log(\sin x)}{\cot x}=\lim_{x\to \pi/2}\frac{\cos x}{\sin x(-\csc^2 x)}=\lim_{x\to \pi/2}\frac{\cos x}{-\sin x}=0$$
$$\Rightarrow \log y\to 0, \text{as}, x\to \pi/2$$
$$\Rightarrow  y\to \exp^0, \text{as}, x\to \pi/2$$
$$\Rightarrow  y\to 1, \text{as}, x\to \pi/2$$
