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Find the equation of the tangent and the normal to the given curve at the indicated point: y = cot2 x – 2 cot x + 2 at x = π/4

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Consider y = cot2 x – 2 cot x + 2 as the equation of curve

Take the value of x = π/4 so we get y = 1

By differentiating both sides w.r.t. x

dy/dx = 2 cot x(- cosec2 x) – 2(- cosec2 x)

By further simplification

dy/dx = – 2 cot x cosec2 x + 2 cosec2 x

We get

(dy/dx)(x = π/4) = 0

Here the equation of tangent at point (π/4, 1)

y – 1 = 0(x – π/4)

So we get

y – 1 = 0

Here y = 1

Here the equation of normal at point (π/4, 1)

y – π/4 = 1/0(x – π/4)

We get

0 = x – π/4

Here x = π/4

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