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If $f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$, for $x \neq \pi$ is continuous at $x=\pi$, then $f(\pi)=$
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$-1$
(A)
$f$ is continuous at $x=\pi$
$\begin{aligned} f(\pi) &=\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{1-\sin x+\cos x}{1+\sin x+\cos x} \\ &=\lim _{x \rightarrow \pi} \frac{-\cos x-\sin x}{\cos x-\sin x} \\ &=\frac{-\cos \pi-\sin \pi}{\cos \pi-\sin \pi}=\frac{-(-1)-0}{-1-0}=\frac{1}{-1}=-1 \end{aligned}$
$f$ is continuous at $x=\pi$
$\begin{aligned} f(\pi) &=\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{1-\sin x+\cos x}{1+\sin x+\cos x} \\ &=\lim _{x \rightarrow \pi} \frac{-\cos x-\sin x}{\cos x-\sin x} \\ &=\frac{-\cos \pi-\sin \pi}{\cos \pi-\sin \pi}=\frac{-(-1)-0}{-1-0}=\frac{1}{-1}=-1 \end{aligned}$
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