Question

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

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

Answer 1

The correct option is B $x$

$\frac{2\sin \alpha }{1+\cos \alpha +\sin \alpha }=x$

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

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

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

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

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

$\Rightarrow$ $\frac{2\sin \alpha (1-\cos \alpha +\sin \alpha )}{2\sin \alpha (2\sin \alpha +1)}=x$

$\Rightarrow$ $\frac{1-\cos \alpha +\sin \alpha }{\sin \alpha +1}=x$

$\therefore$ $\frac{1-\cos \alpha +\sin \alpha }{1+\sin \alpha }=x$