Class 12 Maths Chapter 2: Inverse Trigonometric Functions worksheet with answer key

Answer: (a) −π/6

Principal values of cos⁻¹ lie in [0, π]. 0 = cos⁻¹ 1, π = cos⁻¹(−1) and 5π/6 = cos⁻¹(−√3/2) are attained; −π/6 lies outside [0, π].

Answer: (c) −√3

5π/6 lies in (0, π), so x = cot(5π/6) = −cot(π/6) = −√3.

Answer: (c) (−∞, −1] ∪ [0, ∞)

sec⁻¹ is defined on R − (−1, 1): 2x + 1 ≤ −1 or 2x + 1 ≥ 1, so x ≤ −1 or x ≥ 0. The end points are included, since sec⁻¹(−1) = π and sec⁻¹(1) = 0.

Answer: (b) [1/3, 1]

sin⁻¹ is defined on [−1, 1]: −1 ≤ 3x − 2 ≤ 1 gives 1 ≤ 3x ≤ 3, that is 1/3 ≤ x ≤ 1.

Answer: (b) −π/3

x = cos(5π/6) = −√3/2. sin⁻¹(−√3/2) must lie in [−π/2, π/2], and sin(−π/3) = −√3/2, so sin⁻¹ x = −π/3.

π/2 < 2 < π, and tan 2 = tan(2 − π) with 2 − π ∈ (−π/2, π/2), so tan⁻¹(tan 2) = 2 − π. 3π/2 < 5 < 2π, and cos 5 = cos(2π − 5) with 2π − 5 ∈ [0, π], so cos⁻¹(cos 5) = 2π − 5. Sum = (2 − π) + (2π − 5) = π − 3.

• tan⁻¹(tan 2) = 2 − π, with the branch stated (1)
• cos⁻¹(cos 5) = 2π − 5, and sum π − 3 (1)

Put x = tan θ with θ = tan⁻¹ x ∈ (−π/2, π/2). Then √(1 + x²) = sec θ (positive, since cos θ > 0), so x/√(1 + x²) = tan θ cos θ = sin θ. As θ ∈ [−π/2, π/2], sin⁻¹(sin θ) = θ = tan⁻¹ x.

• substitutes x = tan θ and gets sin θ (1)
• concludes tan⁻¹ x with θ in the branch (1)

An inverse exists only on an interval where the function is one-one and onto. On [−π/2, π/2] cosine is not one-one, since cos(−π/3) = cos(π/3) = 1/2, and it takes no negative values, so it is not onto [−1, 1]. On [0, π] cosine falls steadily from 1 to −1, taking each value in [−1, 1] exactly once, so it is one-one and onto [−1, 1], and cos⁻¹ : [−1, 1] → [0, π] is defined.

• cos is not one-one (or not onto [−1, 1]) on [−π/2, π/2], with an example (1)
• cos is one-one and onto [−1, 1] on [0, π] (1)

(a) sin⁻¹ needs −1 ≤ 1 − x ≤ 1, that is 0 ≤ x ≤ 2: domain [0, 2]. As 1 − x runs over [−1, 1], sin⁻¹(1 − x) runs over [−π/2, π/2], so f runs over [−π, π]: range [−π, π]. (b) 2 sin⁻¹(1 − x) = π/3 gives sin⁻¹(1 − x) = π/6, which lies in [−π/2, π/2], so 1 − x = sin(π/6) = 1/2 and x = 1/2, which is in the domain.

• domain [0, 2] (1)
• range [−π, π] (1)
• x = 1/2 (1)

Put x = cos θ, θ = cos⁻¹ x ∈ [0, π]. Then 2x² − 1 = 2 cos² θ − 1 = cos 2θ. If 0 ≤ x ≤ 1, θ ∈ [0, π/2], so 2θ ∈ [0, π] and cos⁻¹(cos 2θ) = 2θ = 2 cos⁻¹ x. If −1 ≤ x ≤ 0, θ ∈ [π/2, π], so 2θ ∈ [π, 2π], outside the branch; but cos 2θ = cos(2π − 2θ) with 2π − 2θ ∈ [0, π], so cos⁻¹(cos 2θ) = 2π − 2θ = 2π − 2 cos⁻¹ x.

• x = cos θ gives cos 2θ (1)
• case 0 ≤ x ≤ 1: 2θ in [0, π], result 2 cos⁻¹ x (1)
• case −1 ≤ x ≤ 0: uses 2π − 2θ in [0, π] (1)

Let α = tan⁻¹(1/3). sin 2α = 2 tan α/(1 + tan² α) = (2/3)/(10/9) = 3/5. Let β = tan⁻¹(2√2) ∈ (0, π/2), so tan β = 2√2, sec² β = 1 + 8 = 9 and cos β = 1/3 (positive in (0, π/2)). Sum = 3/5 + 1/3 = 14/15.

• sin(2 tan⁻¹(1/3)) = 3/5 (1)
• cos(tan⁻¹ 2√2) = 1/3 with the sign justified (1)
• sum 14/15 (1)