Question

cosy = xcos (a+y)
prove that
dy/dx=cos^2(a+y)\sina​

Answer 1

Answer:

Cosy=xcos(a+y)

x=cosy/cos(a+y)

x=cosysec(a+y)

Differentiating with respect to y

dx/dy =D[cosysec(a+y)]

dx/dy=sec(a+y)D(cosy)+cosyD(sec(a+y))

D(cosy)=-siny

D(sec(a+y))=sec(a+y)tan(a+y)

dx/dy=-sinysec(a+y)+cosysec(a+y)tan(a+y)

dx/dy=sec(a+y)[cosytan(a+y)-siny]

But tan(a+y)=(sin(a+y))/(cos(a+y))

dx/dy=sec(a+y)[cosy [(sin(a+y)/cos(a+y)]-siny]

dx/dy=sec(a+y){[cosysin(a+y)-sinycos(a+y)]/(cos(a+y)}

Sin(A-B)=sinAcosB-cosAsinB

Secx=1/cosx

dx/dy=sec²(a+y)(sin(a+y-y))

dx/dy=sec²(a+y)sina

Taking reciprocal

dy/dx=1/(sec²(a+y)sinA)

1/secx=cosx

dy/dx =cos²(a+y)/sina