Cosy = xcos(a+y) show that Dy/dx …

Given cos y =x cos(a+y)-------(1)

Differentiating both sides w.r.t. x we get

-siny{tex}{dy\over dx}{/tex}= x {tex} { d (cos(a+y)) \over dx}{/tex}+1× cos(a+y)     [ by using product rule and chain rule]

-siny {tex}{dy\over dx}{/tex}=x (-sin(a+y) {tex}{d(a+y) \over dx}{/tex}) +cos(a+y)

-siny {tex}{dy \over dx}{/tex}= x(-sin(a+y)(0+{tex}{dy\over dx}{/tex}))+cos(a+y)

-siny {tex}{dy\over dx}{/tex}=-x sin(a+y) {tex}{dy \over dx}{/tex}+cos(a+y)

-siny {tex}{dy \over dx}{/tex}+x sin(a+y) {tex}{dy \over dx}{/tex}=cos(a+y)

(-siny +x sin(a+y)) {tex}{dy \over dx}{/tex}= cos(a+y)

(-siny + {tex}{Cos y \over cos(a+y)}{/tex}sin(a+y)){tex}{dy\over dx}{/tex}=cos(a+y) [ by eq.(1) substitute the value of x]                         

{tex}{-siny ×cos(a+y)+cosy ×sin(a+y)\over cos(a+y)}{/tex}){tex}{dy\over dx}{/tex}=cos(a+y)

({tex}{sin(a+y)×cos y -cos(a+y)×siny \over cos(a+y)}{/tex}){tex}{dy \over dx}{/tex}=cos(a+y)

({tex}{sin(a+y-y) \over cos(a+y)}{/tex}) {tex}{dy\over dx}{/tex}=cos(a+y)  [ by using identity sin(A-B)=sinAcosB-cosAsinB]

({tex}{sin a\over cos(a+y)}{/tex}){tex}{dy\over dx}{/tex}=cos(a+y)

{tex}{dy \over dx}{/tex}={tex}{cos^2(a+y) \over sin a}{/tex}                Hence Proved.