Question

tan^-1\(\sqrt{x}\) = \(\frac { 1 }{ 2 }\)cos^-1\(\frac { 1-x }{ 1+x }\), x ∈ [0, 1]
 

Answer 1

Put x = tan²θ. ∴ θ = tan^-1\(\sqrt{x}\)
R.H.S.
= \(\frac { 1 }{ 2 }\)cos^-1\(\frac { 1-x }{ 1+x }\)
= \(\frac { 1 }{ 2 }\)cos^-1\(\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)\)
= \(\frac { 1 }{ 2 }\)cos^-1(cos2θ) = \(\frac { 1 }{ 2 }\) x 2θ = θ.
= tan^-1\(\sqrt{x}\) = L.H.S
Hence, tan^-1\(\sqrt{x}\) \(\frac { 1 }{ 2 }\)cos^-1\(\frac { 1-x }{ 1+x }\)