Question

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

Answer 1

It is given that $cosy=xcos(a+y)$

$\frac{d}{dx}\left[cosy\right]=\frac{d}{dx}\left[xcos(a+y)\right]$

$-siny\frac{dy}{dx}=cos(a+y)\ast \frac{d}{dx}\left(x\right)\ast x\frac{d}{dx}\left[cos(a+y)\right]$

$-siny\frac{dy}{dx}=cos(a+y)+x[-sin(a+y)]\frac{d}{dx}$

$[xsin(a+y)-siny]\frac{dy}{dx}=cos(a+y)⟹\left(1\right)$

Since $cosy=xcos(a+y),x=\frac{cosy}{cos(a+y)}$

The eqa (1) reduces to

$[\frac{cosy}{cos(a+y)}\ast sin(a+y)-siny]\frac{dy}{dx}=cos(a+y)$

$[cosy\ast sin(a+y)-siny\ast cos(a+y)]\frac{dy}{dx}={cos}^2(a+y)$

$sin[a+y-y]\frac{dy}{dx}={cos}^2(a+b)$\\ $\frac{dy}{dx}=\frac{{cos}^2(a+b)}{sina}$