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
    $1 + y$

Answer

Correct option: B.
$y$
b
(b) We have, $\frac{{2\sin \alpha }}{{1 + \cos \alpha + \sin \alpha }} = y$

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

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

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free