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1. parametric

x= sec theta y=ln(1+ cos2theta)

find an equation of the tangent to the curve at the point where theta= pi/3

right and ive strugelled finding the dydx with the ln bit and th cos
2. (Original post by ruth_lou)
parametric

x= sec theta y=ln(1+ cos2theta)

find an equation of the tangent to the curve at the point where theta= pi/3

right and ive strugelled finding the dydx with the ln bit and th cos
x = sect
dx/dt = sect.tant

y = ln(1+cos2t)

y = lnu
dy/dt = dy/du . du/dt
dy/dt = (1/u)(-2sin2t)
dy/dt = (-2sin2t)/(1+cos2t)

dy/dx = dy/dt . dt/dx
dy/dx = [(-2sin2t)/(1+cos2t)](1/sect.tant)
dy/dx = (-2sin2t.cost)/[(1+cos2t)tant]

-2sin{2pi/3} = -√3
1+cos{2pi/3} = 1/2
cos{pi/3} = 1/2
tan{pi/3} = √3

dy/dx = -1

.. I think
3. (t=theta)
dy/dx=(dy/dt)/(dx/dt)
dy/dt=dy/du.du/dt ...........(u=1+cos2t)
dy/dt=-2sin2t/(1+cos2t)

dx/dt=secttant

dy/dx=-2sin2t/(1+cos2t)secttant
f '(pi/3)=-1

y-b=f '(a)(x-a)
y+ln2 = -1(x-pi/3)
y+ln2=pi/3-x

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