An equivalent of e^(sin x) − e^(tan x)

Expansion of the difference of the arguments. $\tan x=x+\frac{x^3}3+o(x^3)$ and $\sin x=x-\frac{x^3}6+o(x^3)$, so
$$\tan x-\sin x=\frac{x^3}2+o(x^3).$$

Factoring. Since $\tan x-\sin x\to0$, $1-e^{\tan x-\sin x}\sim-(\tan x-\sin x)\sim-\frac{x^3}2$. Also $e^{\sin x}\to1$. Hence
$$e^{\sin x}-e^{\tan x}\sim-\frac{x^3}2.$$

Warning. Expanding each term separately to order $2$ only gives $0$: both functions are $1+x+\frac{x^2}2+\cdots$. The difference starts at order $3$, which is why the expansions must go that far.

Remark. In general, $e^u-e^v\sim e^v(u-v)$ when $u-v\to0$: it reduces an exponential difference to a simple difference.