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Question: If\(\frac{2\sin\alpha}{\{ 1 + \cos\alpha + \sin\alpha\}} = y,\) then \(\frac{\{ 1 - \cos\alpha + \si...

If2sin⁡α{1+cos⁡α+sin⁡α}=y,\frac{2\sin\alpha}{\{ 1 + \cos\alpha + \sin\alpha\}} = y, then {1−cos⁡α+sin⁡α}1+sin⁡α=\frac{\{ 1 - \cos\alpha + \sin\alpha\}}{1 + \sin\alpha} =

A

1y\frac{1}{y}

B

yy

C

1−y1 - y

D

1+y1 + y

Answer

yy

Explanation

Solution

We have, 2sin⁡α1+cos⁡α+sin⁡α=y\frac{2\sin\alpha}{1 + \cos\alpha + \sin\alpha} = y

Then 4sin⁡α2cos⁡α22cos⁡2α2+2sin⁡α2cos⁡α2=y\frac{4\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\cos^{2}\frac{\alpha}{2} + 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}} = y

⇒ 2sin⁡α2cos⁡α2+sin⁡α2×(sin⁡α2+cos⁡α2)(sin⁡α2+cos⁡α2)=y\frac{2\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2}} \times \frac{\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)}{\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)} = y

⇒ 1−cos⁡α+sin⁡α1+sin⁡α=y\frac{1 - \cos\alpha + \sin\alpha}{1 + \sin\alpha} = y.