(b) यहाँ, $\frac{{2\sin \alpha }}{{1 + \cos \alpha + \sin \alpha }} = y$

तब  $\frac{{4\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{2{{\cos }^2}\frac{\alpha }{2} + 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}} = y$

==> $\frac{{2\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2} + \sin \frac{\alpha }{2}}} \times \frac{{\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)}}{{\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)}} = y$

==> $\frac{{1 – \cos \alpha + \sin \alpha }}{{1 + \sin \alpha }} = y$.