Question

tan⁻¹(√1+x²+√1-x²/√1+x² -√1-x²) = π/4 + 1/2cos⁻¹x²,-1 < x< 1,x≠0,prove it

Answer 1

we have to prove that,

tan⁻¹(√1+x²+√1-x²/√1+x² -√1-x²) = π/4 + 1/2cos⁻¹x², where -1 < x < 1 , x ≠ 0

Let π/4 + 1/2cos⁻¹x² = A .....(1)

⇒A - π/4 = 1/2cos⁻¹x²

⇒2(A - π/4) = 2A - π/2 = cos⁻¹x²

⇒cos(2A - π/2) = x²

we know, cos(-θ) = cosθ

so, cos(2A - π/2) = cos(π/2 - 2A)

also cos(π/2 - θ) = sinθ

so, cos(2A - π/2) = cos(π/2 - 2A) = sin2A.

now, sin2A = x²

use formula, sin2θ = 2tanθ/(1 + tan²θ)

so, sin2A = 2tanA/(1 + tan²A)

⇒x² = 2tanA/(1 + tan²A)

⇒x² + x²tan²A = 2tanA

⇒x²tan²A - 2tanA + x² = 0

⇒tanA = {2 ± √(2² - 4x².(x²))}/2x²

= {2 ± 2√(1 - x⁴)}/2x²

taking tanA = {2 + 2√(1 - x⁴)}/2x²

= {(1 + x²) (1 - x²) + 2√(1 - x²).√(1 + x²)}/{√(1 + x²) - √(1 - x²)}{√(1 + x²) + √(1 - x²)}

= [{√(1 + x²) + √(1 - x²)}]²/[{√(1 + x²) - √(1 - x²)}{√(1 + x²) + √(1 - x²)}]

= [√(1 + x²) + √(1 - x²)]/[√(1 + x²) - √(1 - x²)]

hence, tanA = [√(1 + x²) + √(1 - x²)]/[√(1 + x²) - √(1 - x²)]

⇒A = tan^-1 [√(1 + x²) + √(1 - x²)]/[√(1 + x²) - √(1 - x²)] ........(2)

from equations (1) and (2),

tan⁻¹(√1+x²+√1-x²/√1+x² -√1-x²) = π/4 + 1/2cos⁻¹x² [ proved]