If cos y$=$x cos(a + y), with cos a$\ne \pm$ 1, prove that $\cfrac{{dy}}{{dx}} = \cfrac{{{{\cos }^2}(a + y)}}{{\sin a}}$.

We have, cos y $=$x cos (a + y )
$\Rightarrow$ $x = \cfrac{{\cos y}}{{\cos (a + y)}}$

$\Rightarrow$ $\cfrac{{dx}}{{dy}} = \cfrac{{\cos (a + y)( - \sin y) - \cos y( - \sin (a + y))}}{{{{\cos }^2}(a + y)}}$

$= \cfrac{{\cos y\sin (a + y) - \sin y\cos (a + y)}}{{{{\cos }^2}(a + y)}}$

$= \cfrac{{\sin (a + y - y)}}{{{{\cos }^2}(a + y)}} = \cfrac{{\sin a}}{{{{\cos }^2}(a + y)}}$
therefore, $\cfrac{{dy}}{{dx}} = \cfrac{{{{\cos }^2}(a + y)}}{{\sin a}}$