Question

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

Answer 1

Given,

$\cos y=x\cos (a+y)$

$x=\frac{\cos y}{\cos (a+y)}$

differentiating w.r.t x, we get,

$\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(\frac{\cos y}{\cos (a+y)}\right)$

$1=\frac{d}{dy}\left(\frac{\cos y}{\cos (a+y)}\right)\times \frac{dy}{dx}$

$1=\left(\frac{\sin (a+y)\cos y-\cos (a+y)\sin y}{{\cos }^2(a+y)}\right)\times \frac{dy}{dx}$

$1=\left(\frac{\sin \left(\right(a+y)-y)}{{\cos }^2(a+y)}\right)\times \frac{dy}{dx}$

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

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

Hence proved.