cot -1 (√(1 + x2) – x) = π/2 – ½ cot -1 x
Take x = cot θ
Where θ = cot -1 x
Here LHS = cot -1 (√(1 + x2) – x)
By substituting the value of x
= cot -1 (√(1 + cot2 θ) – cot θ)
It can be written as
= cot -1 (cosec θ – cot θ)
So we get
= cot -1 (1/sin θ – cos θ/sin θ)
On further calculation
= cot -1 ((1 – cos θ)/ sin θ)
We get
= cot -1 (2sin2 θ/2 / 2sin θ/2.cos θ/2)
By simplification
= cot -1 (tan θ/2)
Here,
= cot -1 (cot [π/2 – 1/2 θ])
So
= π/2 – 1/2 θ
By substituting the value of θ
= π/2 – 1/2 cot -1 x
= RHS
Hence, it is proved.