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Question

The solution lof dydxx tan(yx)=1

A
c eex22=sin(yx)
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B
c eex22=sin(y+x)
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C
c eex22=sin(yx)2
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D
c eex22=cos(yx)
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Solution

The correct option is A c eex22=sin(yx)
Put y-x = z. Then dydx1=dzdxdydx=1+dzdx
Given dydxx tan(yx)=11+dzdxx tan z=1dzdx=x tan z1tanzdz=x dxcotz dz=x dx
log|sin z|logc=x22log|sin zc|=x22=sinzc=ex22sin(yx)=c ex22
The solution is sin(yx)=cex22, where c is arbitrary constant.

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