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If xsin(a+y)+sinacos(a+y)=0 , prove that...

If `xsin(a+y)+sinacos(a+y)=0` , prove that `(dy)/(dx)=(sin^2(a+y))/(sina)`

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Verified by Experts

Given that,
`xsin(a+y)+sinacos(a+y)=0`
we know that,
`(UV)'=U'V+VU'`
`=>1*sin*(a+y)+xcos(a+y)y'+sina(-sin(a+y))(0+y')=0`
`=>y'[xcos(a+y)-sinasin(a+y)]=-sin(a+y)`
`=>y'=(-sin(a+y))/(xcos(a+y)-sin(sin(a+y)))*sin(a+y)/(sin(a+y)`
`=>y'=(-sin(a+y))/(-sina[cos^2(a+y)+sin^2(a+y)]`
`=>y'=sin^2(a+y)/(sina)`
`=>(dy)/(dx)=(sin^2(a+y))/(sina)`
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RD SHARMA-DIFFERENTIATION-Solved Examples And Exercises
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  3. If xsin(a+y)+sinacos(a+y)=0 , prove that (dy)/(dx)=(sin^2(a+y))/(sina)

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