Consider f x = tan - 1 ⁡ 1 + sin ⁡ x 1 - sin ⁡ x , x ∈ 0 , π 2 . A normal to y = f x at x = π 6 also passes through the point

Question

Consider f x = tan - 1 ⁡ 1 + sin ⁡ x 1 - sin ⁡ x , x ∈ 0 , π 2 . A normal to y = f x at x = π 6 also passes through the point, π 6 , 0, π 4 , 0, 0 , 0, 0 , 2 π 3

MARKS App · Accepted Answer

f x = tan - 1 ⁡ 1 + sin ⁡ x 1 - sin ⁡ x , x ∈ 0 , π 2 f ′ ( x ) = 1 1 + 1 + sin ⁡ x 1 - sin ⁡ x . 1 2 1 + sin ⁡ x 1 - sin ⁡ x - 1 2 . cos ⁡ x 1 - sin ⁡ x + cos ⁡ 1 + sin ⁡ x 1 - sin ⁡ x 2 at x = π 6 f ′ x = 1 2 2 . 1 2 1 3 3 1 4 = 1 2 Slope of tangent = 1 2 So slope of normal = - 2 Also at x = π 6 , y = tan - 1 ⁡ 3 = π 3 . So equation of the Normal would be y - π 3 = - 2 x - π 6 It passes through 0 , 2 π 3

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