Question

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

• A. $\frac{1}{y}$
• B. $y$
• C. $1-y$
• D. $1+y$

Answer 1

Answer: (B)

$y$

$\frac{1-\cos \alpha +\sin \alpha }{1+\sin \alpha }=\frac{1-\cos \alpha +\sin \alpha }{1+\sin \alpha }.\frac{1+\cos \alpha +\sin \alpha }{1+\cos \alpha +\sin \alpha }=\frac{{(1+\sin \alpha )}^2-{\cos }^2\alpha }{(1+\sin \alpha )(1+\cos \alpha +\sin \alpha )}=\frac{1+{\sin }^2\alpha +2\sin \alpha -(1-{\sin }^2\alpha )}{(1+\sin \alpha )(1+\cos \alpha +\sin \alpha )}=\frac{2{\sin }^2\alpha +2\sin \alpha }{(1+\sin \alpha )(1+\cos \alpha +\sin \alpha )}=\frac{2\sin \alpha (\sin \alpha +1)}{(1+\sin \alpha )(1+\cos \alpha +\sin \alpha )}=\frac{2\sin \alpha }{1+\cos \alpha +\sin \alpha }=y$