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The equation of the curve passing throug...

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

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Given,
`(dy)/(dx)​=y/x​−cos^2(y/x)`​
Multiply by `x` both side,
` x(dy)/(dx)​=xy/x​−xcos^2(y/x)`
`​ x(dy)/(dx)​=y​−xcos^2(y/x)`
` x(dy)/(dx)​-y​=−xcos^2(y/x)`
`(x(dy)-y(dx))/(dx)​​=−xcos^2(y/x)`
`(x(dy)-y(dx))/x​​=−cos^2(y/x)dx`
...
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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)))

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