If $a=\frac{2\sin \theta }{1+\cos \theta +\sin \theta }\text{,}$ then $\frac{1+\sin \theta -\cos \theta }{1+\sin \theta }$ is equal to 

Solution:

We have,

  $a=\frac{2\sin \theta }{1+\cos \theta +\sin \theta }$

$\begin{array}{l}\Rightarrow a=\frac{2\sin \theta (1+\sin \theta -\cos \theta )}{(1+\cos \theta +\sin \theta )(1+\sin \theta -\cos \theta )}\\ \Rightarrow a=\frac{2\sin \theta (1+\sin \theta -\cos \theta )}{(1+\sin \theta {)}^2-{\cos }^2\theta }\\ \Rightarrow a=\frac{2\sin \theta (1+\sin \theta -\cos \theta )}{2\sin \theta (1+\sin \theta )}=\frac{1+\sin \theta -\cos \theta }{1+\sin \theta }\end{array}$