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

We have,

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

On differentiating w,r,t $x$, we get

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

$⟹\frac{dy}{dx}=\frac{\cos (a+y)}{x\sin (a+y)-\sin y}$

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

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

Hence, proved.