MCQ
If $\frac{{2\sin \alpha }}{{\{ 1 + \cos \alpha + \sin \alpha \} }} = y,$ then $\frac{{\{ 1 - \cos \alpha + \sin \alpha \} }}{{1 + \sin \alpha }} = $
- A$\frac{1}{y}$
- ✓$y$
- C$1 - y$
- D$1 + y$
then, $\frac{{4\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{2{{\cos }^2}\frac{\alpha }{2} + 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}} = y$
==> $\frac{{2\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2} + \sin \frac{\alpha }{2}}} \times \frac{{\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)}}{{\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)}} = y$
==> $\frac{{1 - \cos \alpha + \sin \alpha }}{{1 + \sin \alpha }} = y$.
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