Find the equation of the tangent and the normal to the given curve at

1.	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
Answer» 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

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

Answer»

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