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(D) The interior angle of a regular octagon is given by $\alpha = \frac{(8-2) 	imes 180^{\circ}}{8} = 135^{\circ} = \frac{3\pi}{4}$.We need to evalua

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$2$

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(D) The interior angle of a regular octagon is given by $\alpha = \frac{(8-2) 	imes 180^{\circ}}{8} = 135^{\circ} = \frac{3\pi}{4}$.We need to evaluate $\lim_{	heta 	o (\frac{3\pi}{4})^+} \frac{	an 	heta - 1}{[\sin 	heta + \cos 	heta]}$.As $	heta 	o (\frac{3\pi}{4})^+$,$	heta$ is slightly greater than $\frac{3\pi}{4}$.In this interval,$\sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + \frac{\pi}{4})$.As $	heta 	o (\frac{3\pi}{4})^+$,$	heta + \frac{\pi}{4} 	o \pi^+$,so $\sin(	heta + \frac{\pi}{4})$ is a small negative value.Thus,$\sin 	heta + \cos 	heta$ is a small negative value,specifically in the range $(-1, 0)$.Therefore,$[\sin 	heta + \cos 	heta] = -1$.The limit becomes $\lim_{	heta 	o (\frac{3\pi}{4})^+} \frac{	an 	heta - 1}{-1} = -(	an(\frac{3\pi}{4}) - 1) = -(-1 - 1) = -(-2) = 2$.

If $\alpha$ is the interior angle of a regular octagon,then $\lim_{\theta \to \alpha^+} \frac{\tan \theta - 1}{[\sin \theta + \cos \theta]}$ is equal to (Note: $[k]$ denotes the greatest integer less than or equal to $k$).

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