if Cosy= xCos(a+y), show that dy/dx= cos^{2}(a+y)/sin(a)

cosy=xcos(a+y)

Differentiating both sides wrt 'x'

=>d/dx(cos y) =cos(a+y) dx/dx+ xd/dx [cos(a+y) ] (RHS differentiated on the basis of product rule)

=> -siny*dy/dx= cos (a+y)- xsin(a+y)*dy/dx

=> dy/dx [x sin(a+y)-siny] = cos(a+y)

Now x= (cosy)/cos(a+y)

=> dy/dx {[ sin(a+y).cosy-siny.cos(a+y )] / cos(a+y)}=cos(a+y)

=> dx/dy [ sin(a+y-y )]= cos2(a+y) { using sine formula sin(x-y)= sinx.cosy-siny.cosx}

=>dy/d x= cos2(a+y)/ sin(a)

Hope u got it!!!

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