Trigonometric Identities Question 24
Question: If $ \frac{2\sin \alpha }{{1+\cos \alpha +\sin \alpha }}=y, $ then $ \frac{{1-\cos \alpha +\sin \alpha }}{1+\sin \alpha }= $
[BIT Ranchi 1996; Orissa JEE 2004]
Options:
A) $ \frac{1}{y} $
B) $ y $
C) $ 1-y $
D) $ 1+y $
Show Answer
Answer:
Correct Answer: B
Solution:
We have, $ \frac{2\sin \alpha }{1+\cos \alpha +\sin \alpha }=y $ Then $ \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 $
Þ $ \frac{2\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}}\times \frac{( \sin \frac{\alpha }{2}+\cos \frac{\alpha }{2} )}{( \sin \frac{\alpha }{2}+\cos \frac{\alpha }{2} )}=y $
Þ $ \frac{1-\cos \alpha +\sin \alpha }{1+\sin \alpha }=y $ .
BETA
