If p = frac{2 sin theta}{1 + cos theta + sin | vedclass

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(D) Given: $p = \frac{2 \sin 	heta}{1 + \cos 	heta + \sin 	heta}$ and $q = \frac{\cos 	heta}{1 + \sin 	heta}$.Consider $p + q = \frac{2 \sin 	he

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$q + p = 1$

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(D) Given: $p = \frac{2 \sin 	heta}{1 + \cos 	heta + \sin 	heta}$ and $q = \frac{\cos 	heta}{1 + \sin 	heta}$.Consider $p + q = \frac{2 \sin 	heta}{1 + \sin 	heta + \cos 	heta} + \frac{\cos 	heta}{1 + \sin 	heta}$.To simplify $p$,multiply the numerator and denominator by $(1 + \sin 	heta - \cos 	heta)$:$p = \frac{2 \sin 	heta (1 + \sin 	heta - \cos 	heta)}{(1 + \sin 	heta)^2 - \cos^2 	heta} = \frac{2 \sin 	heta (1 + \sin 	heta - \cos 	heta)}{1 + 2 \sin 	heta + \sin^2 	heta - \cos^2 	heta}$.Using $\cos^2 	heta = 1 - \sin^2 	heta$:$p = \frac{2 \sin 	heta (1 + \sin 	heta - \cos 	heta)}{1 + 2 \sin 	heta + \sin^2 	heta - (1 - \sin^2 	heta)} = \frac{2 \sin 	heta (1 + \sin 	heta - \cos 	heta)}{2 \sin 	heta + 2 \sin^2 	heta} = \frac{2 \sin 	heta (1 + \sin 	heta - \cos 	heta)}{2 \sin 	heta (1 + \sin 	heta)} = \frac{1 + \sin 	heta - \cos 	heta}{1 + \sin 	heta} = 1 - \frac{\cos 	heta}{1 + \sin 	heta}$.Since $q = \frac{\cos 	heta}{1 + \sin 	heta}$,we have $p = 1 - q$.Therefore,$p + q = 1$.

If $p = \frac{2 \sin \theta}{1 + \cos \theta + \sin \theta}$ and $q = \frac{\cos \theta}{1 + \sin \theta}$,then

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