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If x cos(a+y)= cosy then prove that `dy/dx=(cos^2(a+y)/sina)`

Hence show that `sina(d^2y)/(dx^2)+sin2(a+y)(dy)/dx=0`

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#### Solution

Given that

x cos(a+y)=cosy...1

`=>x=(cosy)/cos(a+y)`

Differentiating both sides of the equation (1), we have,

`x xx(-sin(a+y))(dy)/(dx)+1xxcos(a+y)=-siny(dy)/dx`

`=>[siny-xsin(a+y)](dy)/dx=-cos(a+y)`

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

`=>[(cos(a+y)xxsiny-cosysin(a+y))/cos(a+y)]dx/dy=-cos(a+y)`

`=>[sin(a+y-y)]dy/dx=-cos^2(a+y) `

`=>[sina]dy/dx=-cos^2(a+y)`

`=>dy/dx=((-cos^2(a+y))/sina) `

Differentiating once again with respect to x, we have,

`sina(d^2y)/dx^2=-2cos(a+y)sin(a+y)dy/dx`

`=>sina((d^2y)/dx^2)+2cos(a+y)sin(a+y)dy/dx=0`

`=>sina(d^2y)/dx^2+sin2(a+y)dy/dx=0`

Hence proved.

Concept: Second Order Derivative

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