MCQ
If $\sin y = x\sin (a + y),$ then ${{dy} \over {dx}} = $
  • A
    ${{{\sin }^2}(a + y)}$
  • B
    ${{{{\sin }^2}(a + y)} \over {\sin (a + 2y)}}$
  • ${{{{\sin }^2}(a + y)} \over {\sin a}}$
  • D
    ${{{{\sin }^2}(a + y)} \over {\cos a}}$

Answer

Correct option: C.
${{{{\sin }^2}(a + y)} \over {\sin a}}$
c
(c) $\sin y = x\sin (a + y)$==>$x = \frac{{\sin y}}{{\sin (a + y)}}$

==> $1 = \frac{{\cos y.\frac{{dy}}{{dx}}.\sin (a + y) - \sin y\cos (a + y)\frac{{dy}}{{dx}}}}{{{{\sin }^2}(a + y)}}$

$ = \frac{{\frac{{dy}}{{dx}}.\sin (a + y - y)}}{{{{\sin }^2}(a + y)}} $

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{{\sin }^2}(a + y)}}{{\sin a}}$.

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