If x cos theta + y sin theta = p is the norma | vedclass

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(D) Given the line equation: $x + 2y + 1 = 0$. Converting to slope-intercept form $y = mx + c$: $2y = -x - 1 \Rightarrow y = -\frac{1}{2}x - \frac{1}{

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$\frac{\pi}{4}$

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(D) Given the line equation: $x + 2y + 1 = 0$. Converting to slope-intercept form $y = mx + c$: $2y = -x - 1 \Rightarrow y = -\frac{1}{2}x - \frac{1}{2}$. Thus,$m = -\frac{1}{2}$ and $c = -\frac{1}{2}$. Converting to normal form $x \cos 	heta + y \sin 	heta = p$: Dividing $x + 2y = -1$ by $\sqrt{1^2 + 2^2} = \sqrt{5}$,we get $\frac{1}{\sqrt{5}}x + \frac{2}{\sqrt{5}}y = -\frac{1}{\sqrt{5}}$. Since the normal form requires $p > 0$,we multiply by $-1$: $-\frac{1}{\sqrt{5}}x - \frac{2}{\sqrt{5}}y = \frac{1}{\sqrt{5}}$. Thus,$\cos 	heta = -\frac{1}{\sqrt{5}}$ and $\sin 	heta = -\frac{2}{\sqrt{5}}$. Therefore,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{-2/\sqrt{5}}{-1/\sqrt{5}} = 2$. Now,calculate $	an^{-1}(	an 	heta + m + c) = 	an^{-1}(2 - \frac{1}{2} - \frac{1}{2}) = 	an^{-1}(1) = \frac{\pi}{4}$.

If $x \cos \theta + y \sin \theta = p$ is the normal form and $y = mx + c$ is the slope-intercept form of the line $x + 2y + 1 = 0$,then $\tan^{-1}(\tan \theta + m + c) = $

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