find the equation of the curve passing through the point `(1,pi/4)` and tangent at any point of which makes an angle `tan^-1(y/x-cos^2(y/x))` with x-axes.

The equation of the curve passing through the point (1,pi/4) and having a slope of tangent at any point (x,y) as y/x - cos^2(y/x) is

The equation of the curve passing through the point (1,pi/4) and having a slope of tangent at any point (x,y) as y/x - cos^2(y/x) is

The equation of the curve passing through the point (1,(pi)/(4)) and having a slope of tangent at any point (x,y) as (y)/(x)-cos^(2)((y)/(x)) is

Find the equation of a curve passing through the point (1,pi/4) if the slope of the tangent to the curve at any point P(x,y) is y/x-cos^2(y/x)

2.The equation of the curve passing through the point (1 (pi)/(4)) and having slope of tangent at any point (x ,y) as (y)/(x)-cos^(2)((y)/(x)) is 1) x=e^(1-tan(y/x)) 2) x=e^(1+tan(y/x)) 3) x=1-tan(y/x) 4) y=e^(1-cot((y)/(x)))

2.The equation of the curve passing through the point (1 (pi)/(4)) and having slope of tangent at any point (x y) as (y)/(x)-cos^(2)(y)/(x) is 1) x=e^(1-tan(y/x)) 2) x=e^((1+tan(y/x)) 3) x=1-tan(y/x)4)y=e^(1-cot((y)/(x)))

Find the equation of a curve passing through (1,1) and whose slope of tangent at a point (x, y) is -(x)/(y) .

Find the equation of a curve passing through (1,1) and whose slope of tangent at a point (x, y) is -(x)/(y) .