Question

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

Answer 1

L.H.S. `tan^-1 [(sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2 - sqrt(1 - x^2)))]`

Put x² = cos θ

∴ θ = `cos^-1 x^2`

⇒ `tan^-1 [(sqrt(1 + cos theta) + sqrt(1 - cos theta))/(sqrt(1 + cos theta) - sqrt(1 - cos theta))]`

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

⇒ `tan^-1 [(cos  theta/2 + sin  theta/2),(cos  theta/2 - sin  theta/2)]`

⇒ `tan^-1 [(1 + tan  theta/2),(1 - tan  theta/2)]`  ......[Dividing the Nr. and Den. by cos θ/2]

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

⇒ `pi/4 + theta/2`

⇒ `pi/4 + 1/2 cos^-1 x^2` R.H.S. ......[Putting θ = cos^–1x²]

Hence proved.