If frac{2sin alpha}{1 + cos alpha + sin alpha | vedclass

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

Question

(B) Given,$\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$. Using half-angle formulas,$\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}

Vedclass · Accepted Answer

$y$

Vedclass · Answer

(B) Given,$\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$. Using half-angle formulas,$\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$,$1 + \cos \alpha = 2\cos^2 \frac{\alpha}{2}$. Substituting these,we get $\frac{2(2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2})}{2\cos^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = y$. Dividing numerator and denominator by $2\cos \frac{\alpha}{2}$,we get $\frac{2\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2}} = y$. Now,consider the expression $E = \frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha}$. Using $1 - \cos \alpha = 2\sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$,the numerator becomes $2\sin^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2} = 2\sin \frac{\alpha}{2} (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})$. The denominator $1 + \sin \alpha = (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2$. Thus,$E = \frac{2\sin \frac{\alpha}{2} (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})}{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = \frac{2\sin \frac{\alpha}{2}}{\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}} = y$.

If $\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y,$ then $\frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha} = $

Similar Questions

$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $

The value of $6(\sin^6 \theta + \cos^6 \theta) - 9(\sin^4 \theta + \cos^4 \theta) + 4$ is

If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\tan (\alpha+2 \beta)$ is equal to

The value of $\frac{4 \sin 9^{\circ} \sin 21^{\circ} \sin 39^{\circ} \sin 51^{\circ} \sin 69^{\circ} \sin 81^{\circ}}{\sin 54^{\circ}}$ is equal to:

Given that $\pi < \alpha < \frac{3\pi}{2},$ then the expression $\sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\cos^2 \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module