Question

If cosy=xcos(a+y) with cosa≠±1,prove that $\frac{dy}{dx}$=$\frac{cos^2(a+y)}{sin\,a}$

Answer 1

It is given that,cosy=xcos(a+y)
∴$\frac{d}{dx}$[cosy]=$\frac{d}{dx}$[xcos(a+y)]
⇒-siny$\frac{dy}{dx}$=cos(a+y).$\frac{d}{dx}$(x)+x.$\frac{d}{dx}$[cos(a+y)]
⇒-siny$\frac{dy}{dx}$=cos(a+y)+x.[-sin(a+y)]$\frac{dy}{dx}$
⇒[xsin(a+y)-siny]$\frac{dy}{dx}$=cos(a+y) ....(1)
Since cosy=xcos(a+y).x=$\frac{cos\,y}{cos(a+y)}$

Then, equation (1) reduces to
[$\frac{cos\,y}{cos(a+y)}$.sin(a+y)-siny]$\frac{dy}{dx}$cos(a+y)
⇒[cosy.sin(a+y)-siny.cos(a+y)].$\frac{dy}{dx}$=cos2(a+y)
⇒sin(a+y-y)$\frac{dy}{dx}$=cos2(a+b)
⇒ $\frac{dy}{dx}$=$\frac{cos^2(a+b)}{sin\,a}$
Hence, it proved.