The function f(x) = frac{1 - sin x + cos x}{1 | vedclass

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(C) To make $f(x)$ continuous at $x = \pi$,we must find the limit $\lim_{x 	o \pi} f(x)$.Given $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x

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$-1$

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(C) To make $f(x)$ continuous at $x = \pi$,we must find the limit $\lim_{x 	o \pi} f(x)$.Given $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$.Using trigonometric identities $1 + \cos x = 2\cos^2(\frac{x}{2})$ and $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$:$f(x) = \frac{2\cos^2(\frac{x}{2}) - 2\sin(\frac{x}{2})\cos(\frac{x}{2})}{2\cos^2(\frac{x}{2}) + 2\sin(\frac{x}{2})\cos(\frac{x}{2})}$Dividing numerator and denominator by $2\cos(\frac{x}{2})$:$f(x) = \frac{\cos(\frac{x}{2}) - \sin(\frac{x}{2})}{\cos(\frac{x}{2}) + \sin(\frac{x}{2})}$Dividing by $\cos(\frac{x}{2})$:$f(x) = \frac{1 - 	an(\frac{x}{2})}{1 + 	an(\frac{x}{2})} = 	an(\frac{\pi}{4} - \frac{x}{2})$Now,$\lim_{x 	o \pi} f(x) = 	an(\frac{\pi}{4} - \frac{\pi}{2}) = 	an(-\frac{\pi}{4}) = -1$.Thus,$f(\pi) = -1$.

The function $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ is not defined at $x = \pi$. The value of $f(\pi)$,so that $f(x)$ is continuous at $x = \pi$,is

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