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

HINT
$$\lim_{x \to a}f(x)^{g(x)} = \left(\lim_{x \to a} f(x) \right)^{\lim_{x \to a}g(x)}$$ if $f(x)$ and $g(x)$ are continuous in the neighborhood of $a$ and $\lim_{x \to a} f(x) \in \mathbb{R}, \lim_{x \to a}g(x) \in \mathbb{R}$. This is all you need to compute the limit you want.

This is the answer to the new question.
\begin{align}
\underbrace{\lim_{x \to \pi/2} (\sin(x))^{\tan(x)}  = \lim_{t \to 0} \cos(t)^{\cot(t)}}_{t \mapsto \pi/2-x} &= \lim_{t \to 0} \left(1 - 2 \sin^2(t/2)\right)^{\cot(t)}\\
&= \lim_{t \to 0} \left(1 - 2 \sin^2(t/2)\right)^{\frac1{2\sin^2(t/2)} \cdot 2 \sin^2(t/2)\cot(t)}
\end{align}
Now $$2 \sin^2(t/2)\cot(t) =2\sin^2(t/2) \dfrac{\cos(t)}{\sin(t)} = 2\sin^2(t/2) \dfrac{\cos(t)}{2 \sin(t/2) \cos(t/2)} = \cos(t) \tan(t/2)$$
Now let $f(t) = \left(1 - 2 \sin^2(t/2)\right)^{\frac1{2\sin^2(t/2)}}$, $g(t) = \cos(t) \tan(t/2)$ and $a=0$.
You might also make use of the following limit
$$\lim_{y \to 0}(1-y)^{1/y} = \dfrac1e$$ to evaluate $\lim_{t \to 0} f(t)$.
