Prove that ((√1 + sin x + √1 - sin x)/(√1 + sin x - √1 - sin x)) = x/2, x ∈ (0, π/4)

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 us proceed from LHS:

[(√1 + sin x + √1 - sin x)/(√1 + sin x - √1 - sin x)]

= [(√1 + sin x + √1 - sin x)² / (√1 + sin x)² - (√1 - sin x)²] (by rationalizing )

= [(1 + sin x) + (1 - sin x) + 2 √(1 + sin x)(1 - sin x)] / [(1 + sin x - 1 + sin x)]

Using trigonometric identities,

= 2 (1 + √1 - sin² x) / (2 sin x)

= (1 + cos x) / sin x

= (2 cos² x/2) / (2 sin x/2 cos x/2)

= cot x/2

Thus,

LHS

= cot^{- 1} ((√1 + sin x + √1 - sin x) / (√1 + sin x - √1 - sin x))

= cot^{- 1} (cot x/2)

= x/2

= RHS

Hence LHS = RHS

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 10

Prove that ((√1 + sin x + √1 - sin x)/(√1 + sin x - √1 - sin x)) = x/2, x ∈ (0, π/4)

Summary:

Hence we have proved by using inverse trigonometric functions that ((√1 + sin x + √1 - sin x)/(√1 + sin x - √1 - sin x)) = x/2, x ∈ (0, π/4)