Csdrhrt

Here in this answer I have tried to perform partial fraction for $frac{1}{sin(x-a)sin(x-b)}$ as follows:
$$frac{1}{sin(x-a)sin(x-b)}=frac{p}{sin(x-a)}+frac{q}{sin(x-b)} quad(1)$$
Multiplying $(1)$ by $sin(x-a)$ and plugging $x=a$ yields:
$$left(frac{1}{sin(x-b)}=p+qunderset{rightarrow 0}{frac{sin(x-a)}{sin(x-b)}}right)bigg|_{x=a}Rightarrow p=frac{1}{sin(a-b)}$$
Also to obtain q, multiply $(1)$ by $sin(x-b)$ and plugg $x=b$ gives:
$$left(frac{1}{sin(x-a)}=punderset{rightarrow 0}{frac{sin(x-b)}{sin(x-a)}}+qright)bigg|_{x=b}Rightarrow q=frac{1}{sin(b-a)}=-p$$
So that means we can rewrite the original expression as: $$frac{1}{sin(x-a)sin(x-b)}=frac{1}{sinleft(a-bright)}left(frac{1}{sinleft(x-aright)}-frac{1}{sinleft(x-bright)}right)$$
However as pointed out in the comments this doesn't verify. Can anyone point out where I went sideways with this?

Also the same approach can be found here. Is there a rigurous way to prove that we can't perform partial fraction in this case?

Further question, what a about a general case such as proving/disproving: $${prod_{k=1}^nfrac{1}{sin(x-x_k)}}neqsum_{k=1}^n frac{a_k}{sin(x-x_k)}$$

What is really happening is that $xmapstosin(x-x_k)$ is a rational function of $exp(ix)$, so $displaystyleprod_{k=1}^nfrac{1}{sin(x-x_k)}$ has a partial fraction decomposition of the form
begin{align*}
prod_{k=1}^nfrac{1}{sin(x-x_k)}
&=prod_{k=1}^nfrac{2iexp(-ix_k)exp(ix)}{[exp(ix)]^2-[exp(ix_k)]^2}
\
&=sum_{k=1}^nleft(frac{alpha^-_k}{exp(ix)+exp(ix_k)}+frac{alpha^+_k}{exp(ix)-exp(ix_k)}right)
end{align*}
with $alpha_k^pminmathbb{C}$. To be able to say
$$
prod_{k=1}^nfrac{1}{sin(x-x_k)}=sum_{k=1}^n frac{a_k}{sin(x-x_k)}
$$
is thus the condition $alpha_k^+=alpha_k^-$ which we can check holds if and only if $n$ is odd.

The problem is there are no constants $p,,q$ satisfying the intended identity. The reason $frac{1}{(x-a)(x-b)}=frac{p}{x-a}+frac{q}{x-b}$ works is because the denominators are degree-$1$ polynomials, allowing the numerators to be degree-$0$. More generally distinct functions $f,,g$ give $frac{1}{fg}=frac{p}{f}+frac{q}{g}$ with functions $p=frac{1}{g-f},,q=-p$, among other options. But usually there isn't a solution with constant $p,,q$.

We actually have
$$frac1{sin(x-a)sin(x-b)} = frac1{sin(x-b)-sin(x-a)}left(frac1{sin(x-a)}-frac1{sin(x-b)}right)$$
That factor of $sin(x-b)-sin(x-a)$ is equal to $sin(a-b)$ at $a$ and at $b$, but not everywhere; it isn't constant.

Partial fractions are a theorem of linear algebra. If the factors we're dealing with aren't linear or at least polynomial, the theorem isn't valid. There is no partial fraction expansion of the type you're looking for here.