I am very new to Pade' approximation concept, so some detailed derivation for the approximate result of the following function would be very helpful.
The function that I wish to approximate in the limit $x,y\ll1$ is :
$$1-\frac{(1-x^2)\sin^2(\theta)\sin^2(\theta-y)}{(1-(1-x^2)^{1/2}\cos(\theta)\cos(\theta-y))^2}$$
I would like to have an approximation which is correct upto $2^{nd}$ order in both $x,y$.
A single variable problem.
Assuming that you know Taylor series, building the $[m,n]$ Padé approximant of $f(x)$ is quite simple (I shall make it around $x=0$ and there is no problem to generalize).
So, we look for $$f(x)=\frac{\sum_{i= 0}^m a_i\,x^i}{1+\sum_{i= 1}^n b_i \,x^i }$$ and, on the other side, we have the corresponding Taylor expansion $$f(x)=\sum_{i= 0}^p c_i\,x^i \qquad \text{with} \qquad p>m+n$$ So, write $$\Bigg[1+\sum_{i= 1}^n b_i \,x^i\Bigg]\Bigg[\sum_{i= 0}^p c_i\,x^i\Bigg]=\sum_{i= 0}^m a_i\,x^i$$
Now, as usual, you expand the lhs and identify the coefficients.
Example
SImilar to your problem, consider $$f(x)=1-\frac{\alpha \left(1-x^2\right)}{\left(1-\beta \sqrt{1-x^2}\right)^2}$$ The $[2,2]$ Padé approximant is then $$f(x)=\frac{\left(1-\frac{\alpha }{(\beta -1)^2}\right)+\frac{\alpha (3 \beta -4)-3 (\beta -1)^2 \beta }{4 (\beta -1)^3}x^2 } {1+\frac{3 \beta }{4 (\beta -1)} x^2}$$
Now, we shall compare $$\text{Padé - Taylor}=\frac{\alpha (\beta -2) \beta }{16 (\beta -1)^5}x^6+O\left(x^8\right)$$ Do you see the difference ?