If \(f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}\), for \(x \neq \pi\) is continuous at \(x=\pi\), then the value of \(f(\pi)\) is

\(f(x)=\frac{(1+\cos x)-(\sin x)}{(1+\cos x)+(\sin x)}\), is continuous at \(x=\pi\)
\(\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{\left(2 \cos ^2 \frac{x}{2}\right)-\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}{\left(2 \cos ^2 \frac{x}{2}\right)+\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} \)
\( =\lim _{x \rightarrow \pi} \frac{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)} \)
\( =\lim _{x \rightarrow \pi} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\lim _{x \rightarrow \pi} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\) \(\tan \left(\frac{\pi}{4}-\frac{\pi}{2}\right)\) \(=-1\)