The Taylor series expansion of \[\frac{{\sin {\text{x}}}}{{{\text{x}} - \pi }}\]  at \[{\text{x}} = \pi \]  is given by

The correct option is. D. −1+(x−π)23!-..
We want the **Taylor (Maclaurin-type) series expansion** of

$$
f(x) = frac{sin x}{x - pi}, quad text{about } x = pi.
$$

Let

$$
h = x - pi quad Rightarrow quad x = pi + h.
$$

Then

$$
f(x) = frac{sin(pi + h)}{h}.
$$

$$
sin(pi + h) = -sin h.
$$

So

$$
f(x) = frac{-sin h}{h}.
$$
Use series for $sin h$

$$
sin h = h - frac{h^3}{3!} + frac{h^5}{5!} - cdots
$$

So

$$
frac{sin h}{h} = 1 - frac{h^2}{3!} + frac{h^4}{5!} - cdots
$$

Thus

$$
f(x) = -left(1 - frac{h^2}{6} + frac{h^4}{120} - cdots right).
$$

Substitute back $h = x - pi$

$$
f(x) = -1 + frac{(x - pi)^2}{6} - frac{(x - pi)^4}{120} + cdots
$$
(Taylor series of $tfrac{sin x}{x-pi}$ at $x=pi$):**

$$
frac{sin x}{x-pi} = -1 + frac{(x - pi)^2}{6} - frac{(x - pi)^4}{120} + frac{(x - pi)^6}{5040} - cdots
$