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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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 $p + q = \frac{2 \sin 	heta}{1 + \sin 	heta + \cos 	heta} + \frac{\cos 	heta}{1 + \sin 	heta}$.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

Similar Questions

If $x \sin^{2} 60^{\circ} - \frac{3}{2} \sec 60^{\circ} \tan^{2} 30^{\circ} + \frac{4}{5} \sin^{2} 45^{\circ} \tan^{2} 60^{\circ} = 0$,then $x$ is

If $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3},$ the value of $A + B$ is

$\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma ) = $

If $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha,$ then $\cos 2\beta$ is equal to

$\cos 1^\circ \cdot \cos 2^\circ \cdot \cos 3^\circ \dots \cos 179^\circ = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module