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

Step-by-step solution

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 =