The slope of the tangent at `(x , y)` to a curve passing through `(1,pi/4)` is given by `y/x-cos^2(y/x),` then the equation of the curve is :

Options

A. `y=tan^(-1)[log((e)/(x))]`
B. `y=x tan^(-1)[log((x)/(e))]`
C. `y=x tan^(-1)[log((e)/(x))]`
D. None of these

Answer

C

Explanation

According to the given condition, `(dy)/(dx)=(y)/(x)-cos^(2)((y)/(x))`
On putting y=mx
`Rightarrow (dy)/(dx)=b+x(dv)/(dx),` we get
`v+x(dy)/(dx)=v-cos^(2)v`
`Rightarrow (dv)/(cos^(2)v)=-(dx)/(x)`
`Rightarrow sec^(2)vdv=(-1)/(x)dx`
On integrating both sides, we get
`tan v=-log x+log x`
`Rightarrow tan((y)/(x))=-log x+log c`
Since, this curve is pasing through `(1,pi//4)`,
`therefore tan((pi)/(4))=-log1+log cRightarrow log c=1`
`therefore tan ((y)/(x))=-log x+1`
`Rightarrow tan((y)/(x))=-logx+log x`
`Rightarrow y=xtan ^(-1)[log((e)/(x))]`