this one is similar to me last question.
it from May/June 2008 paper 3 which iv attached and its question number 8 part ii.
In the diagram the tangent to a curve at a general point P with coordinates (x, y) meets the x-axis at T.
The point N on the x-axis is such that PN is perpendicular to the x-axis. The curve is such that, for all
values of x in the interval 0 < x < 1/2(Pi), the area of triangle PTN is equal to tan x, where x is in radians.
(i) Using the fact that the gradient of the curve at P is PN/TN, show that
dy/dx = 1/2 y^2 cot x. (i can do part i. i just put it in to help with part ii)
(ii) Given that y = 2 when x = 1/6(pi), solve this differential equation to find the equation of the curve,
expressing y in terms of x.
the diagram is in the attachment
please help
