if Cosy= xCos(a+y), show that dy/dx= cos2(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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