maths --plse help me..

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

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  1. y'(x) = 2cosx+2cos2x
    y'(pi/3) = 1-1 = 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

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  2. plse read the question carefully...its find the area of quadrilateral not the rectangle.

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  3. it happens that the quadrilateral formed by those lines is a rectangle.

    the tangent is horizontal and the normal is vertical.

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