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

Vedclass · Accepted Answer

$q + p = 1$

Vedclass · Answer

(D) Given $p = \frac{2\sin 	heta}{1 + \cos 	heta + \sin 	heta}$ and $q = \frac{\cos 	heta}{1 + \sin 	heta}$.Consider $p + q = \frac{2\sin 	heta}{1 + \sin 	heta + \cos 	heta} + \frac{\cos 	heta}{1 + \sin 	heta}$.$p + q = \frac{2\sin 	heta (1 + \sin 	heta) + \cos 	heta (1 + \sin 	heta + \cos 	heta)}{(1 + \sin 	heta + \cos 	heta)(1 + \sin 	heta)}$.$p + q = \frac{2\sin 	heta + 2\sin^2 	heta + \cos 	heta + \sin 	heta \cos 	heta + \cos^2 	heta}{(1 + \sin 	heta + \cos 	heta)(1 + \sin 	heta)}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have:$p + q = \frac{2\sin 	heta + \sin^2 	heta + 1 + \cos 	heta + \sin 	heta \cos 	heta}{(1 + \sin 	heta + \cos 	heta)(1 + \sin 	heta)}$.$p + q = \frac{\sin^2 	heta + 2\sin 	heta + 1 + \cos 	heta (1 + \sin 	heta)}{(1 + \sin 	heta + \cos 	heta)(1 + \sin 	heta)}$.$p + q = \frac{(1 + \sin 	heta)^2 + \cos 	heta (1 + \sin 	heta)}{(1 + \sin 	heta + \cos 	heta)(1 + \sin 	heta)}$.$p + q = \frac{(1 + \sin 	heta)(1 + \sin 	heta + \cos 	heta)}{(1 + \sin 	heta + \cos 	heta)(1 + \sin 	heta)} = 1$.

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

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