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.