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Question
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.
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