calculus - Find the limit of $lim_{xto 0} (frac{1+tan x}{1+sin x})^{csc^3x}$

I failed to find the limit of:lim(x->0) $(\frac{1+tan(x)}{1+sinx})^{\frac{1}{sin^3(x)}}$?
as X approches 0
How do I find the answer for this?

Thanks in advance. the answer supposed to be sqr(e). but my answer was 1.
Can anyone please help me find my mistake?
I DID:

$lim_{x \to 0} (\frac{1+tan(x)}{1+sin(x)})^{\frac{1}{sin^3(x)}}$

$lim_{x \to 0} (\frac{((1+tan(x))^{1/sin(x)}}{((1+ sin(x))^{1/sin(x)}})^{1/sin^2(x)}$

now I look inside:

$lim_{x \to 0} ((1+tan(x))^{1/sin(x)}$ is e
$lim_{x \to 0} ((1+sin(x))^{1/sin(x)}$ is also e

so we get:

$lim_{x \to 0} (\frac{e}{e})^{\frac{1}{sin^2(x)}}$

$lim_{x \to 0} (1)^{\frac{1}{sin^2(x)}}$ = 1