Question

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

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

Answer 1

The correct option is C $x$

$x=\frac{2sin\alpha }{1+sin\alpha +cos\alpha }=\frac{2sin\alpha }{(1+sin\alpha )+\left(cos\alpha \right)}\times \frac{(1+sin\alpha )-\left(cos\alpha \right)}{(1+sin\alpha )-\left(cos\alpha \right)}=\frac{2sin\alpha (1+sin\alpha -cos\alpha )}{{(1+sin\alpha )}^2-{\left(cos\alpha \right)}^2}=\frac{2sin\alpha (1+sin\alpha -cos\alpha )}{1+{sin}^2\alpha +2sin\alpha -{cos}^2\alpha }=\frac{2sin\alpha (1+sin\alpha -cos\alpha )}{1+{sin}^2\alpha +2sin\alpha -1+{sin}^2\alpha }=\frac{2sin\alpha (1+sin\alpha -cos\alpha )}{2{sin}^2\alpha +2sin\alpha }=\frac{2sin\alpha (1+sin\alpha -cos\alpha )}{2sin\alpha (1+sin\alpha )}=\frac{(1+sin\alpha -cos\alpha )}{(1+sin\alpha )}$