Prove that tan^{- 1} (√1 + x - √1 - x)/(√1 + x + √1 - x) = π/2 - 1/2 cos^{- 1} x, - 1/√2 ≤ x ≤ 1

Solution:

Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios

Here the basic trigonometric function of Sin θ = y, can be changed to θ = sin^-1 y

Let x = cos 2θ

⇒ θ = 1/2 cos^{- 1} x

Thus,

LHS = tan^{- 1} (√1 + x - √1 - x) / (√1 + x + √1 - x)

= tan^{- 1} (√1 + cos 2θ - √1 - cos 2θ) / (√1 + cos 2θ + √1 - cos 2θ)

= tan^{- 1} (√2 cos² θ - √2 sin² θ) / (√2 cos² θ + √2 sin² θ)

= tan^{- 1} (√2 cos θ - √2 sinθ) / (√2 cosθ + √2 sin θ)

= tan^{- 1} [(cos θ - sin θ) / (cos θ + sin θ)]

= tan^{- 1} [(1 - tanθ) / (1 + tan θ)]

= tan^{- 1} (1) + tan^{- 1} (tan θ)

= π/4 - θ

= π/4 - 1/2 cos^{- 1} x

= RHS

Hence, LHS = RHS

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 11

Prove that tan^{- 1} (√1 + x - √1 - x)/(√1 + x + √1 - x) = π/2 - 1/2 cos^{- 1} x, - 1/√2 ≤ x ≤ 1.

Summary:

Hence we have proved by using inverse trigonometric functions that tan^{- 1} (√1 + x - √1 - x)/(√1 + x + √1 - x) = π/2 - 1/2 cos^{- 1} x, - 1/√2 ≤ x ≤ 1