Answer
$$f'(\pi)$$
Work Step by Step
Since
$$f^{\prime}(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\tag{1}$$
and if $f(\theta )= \cos \theta-\sin \theta $, then $f(\pi)=1$
Rewrite the limit as
\begin{aligned}
\lim _{\theta \rightarrow \pi} \frac{\cos \theta-\sin \theta+1}{\theta-\pi} &=\lim _{\theta \rightarrow \pi} \frac{\cos \theta-\sin \theta-(-1)}{\theta-\pi} \\
&=\frac{f(\theta)-f(\pi)}{\theta-\pi},\ \ \ \ \text{from }\ (1)\\
&=f'(\pi)
\end{aligned}
