Question

Prove that `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x))) = pi/4 - 1/2 cos^-1 x, -1/sqrt(2) ≤ x ≤ 1`.

[Hint: Put x = cos 2θ]

Answer 1

L.H.S. = `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`

Put x = cos 2θ

∴ `θ = 1/2 cos^-1x`

∴ L.H.S. = `tan^-1 [(sqrt(1 + cos theta) - sqrt(1 - cos theta))/(sqrt(1 + cos theta) + sqrt(1 - cos theta))]`

= `tan^-1 [(sqrt(2 cos^2(theta/2)) - sqrt(2 sin^2 (theta/2)))/(sqrt(2 cos^2 (theta/2)) + sqrt(2 sin^2 (theta/2)))]`

= `tan^-1 [(sqrt(2) cos (theta/2) - sqrt(2) sin (theta/2))/(sqrt(2) cos (theta/2) + sqrt(2) sin (theta/2))]`

= `tan^-1 [((sqrt(2) cos (theta/2))/(sqrt(2) cos (theta/2)) - (sqrt(2) sin (theta/2))/(sqrt(2) cos (theta/2)))/((sqrt(2) cos (theta/2))/(sqrt(2) cos (theta/2)) + (sqrt(2) sin (theta/2))/(sqrt(2) cos (theta/2)))]`

= `tan^-1 [(1 - tan(theta/2))/(1 + tan (theta/2))]`

= `tan^-1 [(tan  pi/4 - tan (theta/2))/(1 + tan  pi/4. tan (theta/2))]`   ...`[∵ tan  pi/4 = 1]`

= `tan^-1 [tan (pi/4 - theta/2)]`

= `pi/4 - theta/2`

= `pi/4 - 1/2 cos^-1x`   ...[∵ θ = cos^–1x]

= R.H.S.