devarshi-dt-logo

Question:

Consider f(x) = tan⁻¹(√((1+sin x)/(1-sin x))), x ∈ (a, π/2). A normal to y = f(x) at x = π/6 also passes through the point: (0, 0), (0, 2π/3), (π/6, 0), (π/4, 0)

(0, 2π/3)

(π/6, 0)

(π/4, 0)

(0, 0)

Solution:

f(x) = tan⁻¹(√((1+sin x)/(1-sin x))) x ∈ (0, π/2)
Given
Slope at x = π/6, y(π/6) = tan⁻¹(√3) = π/3
f'(x) = 1/(1 + ((1+sin x)/(1-sin x))) * 1/2√((1+sin x)/(1-sin x)) * [(cos x(1-sin x) + cos x(1+sin x))/(1-sin x)²]
f'(π/6) = 1/(1+3) * 1/(2√3) * [(√3/2(1-1/2) + √3/2(1+1/2))/(1-1/2)²] = 1/4 * 1/(2√3) * (√3/4 + (3√3)/4)/(1/4) = 1/4 * 1/(2√3) * (4√3/4) * 4 = 1/2
∴Slope of normal = -2
So, equation of line is y - π/3 / x - π/6 = -2
(0, 2π/3) satisfies this equation -π/3 / -π/6 = 2
So, (0, 2π/3) lies on the line.