Question

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

Answer 1

Given

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

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

Diff w.r.t 'x'

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

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

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

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

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

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

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

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

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