Application of DerivativeMedium

Question

For the curve represented parametrically by the equation, x = 2 ln cot t + 1 and y = tan t + cot t

Options

A.tangent at t = θ/4 is parallel to x-axis
B.tangent at t = θ/4 is parallel to y-axis
C.tangent at t = θ/4 is parallel to the line y = x
D.normal at t = θ/4 is parallel to the line y = x

Solution


at t = = - 4
= sec2 t - cosec2 t
at t = = 0
= 0 for tangent & hence it is parallel to x-axis & its normal is parallel to y axis

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