If $x\sin (a + y) + \sin a \cdot \cos (a + y) = 0,$ then prove that
$\frac{{dy}}{{dx}} = \frac{{{{\sin }^2}(a + y)}}{{\sin a}}$

We have, $x\sin (a + y) + \sin a \cdot \cos (a + y) = 0$
$\Rightarrow$ $x\sin (a + y) = - \sin a \cdot \cos (a + y)$
$\Rightarrow$ $x = \frac{{ - \sin a \cdot \cos (a + y)}}{{\sin (a + y)}}$

$\Rightarrow$ $x = - \sin a \cdot \cot (a + y)$

therefore,$\frac{{dx}}{{dy}} = - \sin a \cdot \left[ { - {{{\mathop{\rm cosec}\nolimits} }^2}(a + y)} \right] \cdot \frac{d}{{dy}}(a + y)$

$= \sin a \cdot \frac{1}{{{{\sin }^2}(a + y)}} \cdot 1$

$= \frac{{{{\sin }^2}(a + y)}}{{\sin a}}$