Question

cot⁻¹(√1+x²-1/x)= π/2 -1/2 tan⁻¹x,prove it

Answer 1

we have to prove that,

cot⁻¹(√1+x²-1/x)= π/2 -1/2 tan⁻¹x

Let π/2 - 1/2 tan⁻¹x = A .......(1)

⇒π/2 - A = 1/2 tan⁻¹x

⇒2(π/2 - A) = tan⁻¹x

⇒π - 2A = tan⁻¹x

⇒tan(π - 2A) = x

we know, tan(π - θ) = -tanθ

so, tan(π - 2A) = -tan2A = x

⇒tan2A = -x

[ tan2A is negative only if 2A lies between π/2 to π or 3π/2 to 2π.

i.e., π/2 < 2A < π ⇒π/4 < A < π/2.

similarly, 3π/2 < 2A < 2π ⇒3π/4 < A < π , hence tanA will be positive and negative both ]

we know, tan2θ= (2tanθ)/(1- tan²θ)

so, tan2A = 2tanA/(1 - tan²A)

⇒-x = 2tanA/(1 - tan²A)

⇒-x + xtan²A = 2tanA

⇒xtan²A - 2tanA - x = 0

⇒tanA = {2 ± √(2² - 4x.(-x))}/2x

⇒tanA = {2 ± 2(1 + x²)}/2x

⇒tanA = {1 ± √(1 + x²)}/x

taking positive value of tanA ,

i.e., tanA = {1 + √(1 + x²)}/x

then, cotA = x/{1 + √(1 + x²)}

after rationalisation,

cotA = {√(1 + x²) - 1}/x

⇒A = cot⁻¹ [ {√(1 + x²) - 1}/x ] .......(2)

from equations (1) and (2),

cot⁻¹(√1+x²-1/x)= π/2 -1/2 tan⁻¹x [ proved]