Prove that: tan⁻¹[√(1+x) - √(1-x) / √(1+x) + √(1-x)] = π/4 - (1/2)cos⁻¹x, √2 ≤ x ≤ 1

Take LHS,
Put x = cos2θ
tan⁻¹[√(1+cos2θ) - √(1-cos2θ) / √(1+cos2θ) + √(1-cos2θ)]
= tan⁻¹[√(1+2cos²θ -1) - √(1 - (2cos²θ -1)) / √(1+2cos²θ -1) + √(1 - (2cos²θ -1))]
= tan⁻¹[√(2cos²θ) - √(2-2cos²θ) / √(2cos²θ) + √(2-2cos²θ)]
= tan⁻¹[√2cosθ - √2sinθ / √2cosθ + √2sinθ]
= tan⁻¹[cosθ - sinθ / cosθ + sinθ]
= tan⁻¹[1 - tanθ / 1 + tanθ]
= tan⁻¹[tan(π/4 - θ)]
= π/4 - θ
as x = cos2θ so, θ = (1/2)cos⁻¹x
= π/4 - (1/2)cos⁻¹x = RHS
Hence proved.