maths plse help me..
posted by Anamika on .
if tangent to the curve y=2sinx +sin2x are drawn at p(x=60degree) then find the area of the quadrilateral formed by the tangent the normal at P and the coordinate axis

y'(x) = 2cosx+2cos2x
y'(pi/3) = 11 = 0
y(pi/3) = √3+√3/2 = 3√3/2
so, we have the axes, and the lines y=3√3/2 and x=pi/3
the rectangle has area pi√3/2 
plse read the question carefully...its find the area of quadrilateral not the rectangle.

it happens that the quadrilateral formed by those lines is a rectangle.
the tangent is horizontal and the normal is vertical.