NCERT Solutions›Class 12 Mathematics›Chapter 2

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Chapter 2 · Class 12 Mathematics

Inverse Trigonometric Functions

3 exercises41 questions solved

Exercise 2.1 Exercise 2.2 Exercise Misc

Exercise 2.1Principal Values of Inverse Trigonometric Functions

Find the principal value of sin⁻¹(–1/2).

Solution

sin⁻¹(–1/2) = –sin⁻¹(1/2) [since sin⁻¹ is an odd function] sin(π/6) = 1/2, so sin⁻¹(1/2) = π/6 ∴ sin⁻¹(–1/2) = –π/6

Find the principal value of cos⁻¹(√3/2).

Solution

We need θ ∈ [0, π] such that cos θ = √3/2. cos(π/6) = √3/2 ∴ cos⁻¹(√3/2) = π/6

Find the principal value of cosec⁻¹(2).

Solution

cosec⁻¹(2) = sin⁻¹(1/2) [since cosec⁻¹x = sin⁻¹(1/x) for |x| ≥ 1] sin(π/6) = 1/2 ∴ cosec⁻¹(2) = π/6

Find the principal value of tan⁻¹(–√3).

Solution

tan⁻¹(–√3) = –tan⁻¹(√3) [tan⁻¹ is an odd function] tan(π/3) = √3, so tan⁻¹(√3) = π/3 ∴ tan⁻¹(–√3) = –π/3

Find the principal value of cos⁻¹(–1/2).

Solution

cos⁻¹(–x) = π – cos⁻¹(x) [property of cos⁻¹] cos⁻¹(1/2) = π/3 [since cos(π/3) = 1/2] ∴ cos⁻¹(–1/2) = π – π/3 = 2π/3

Find the principal value of tan⁻¹(–1).

Solution

tan⁻¹(–1) = –tan⁻¹(1) tan(π/4) = 1, so tan⁻¹(1) = π/4 ∴ tan⁻¹(–1) = –π/4

Find the principal value of sec⁻¹(2/√3).

Solution

sec⁻¹(2/√3) = cos⁻¹(√3/2) [since sec⁻¹x = cos⁻¹(1/x)] cos(π/6) = √3/2 ∴ sec⁻¹(2/√3) = π/6

Find the principal value of cot⁻¹(√3).

Solution

cot⁻¹(√3) = tan⁻¹(1/√3) [since cot⁻¹x = tan⁻¹(1/x) for x > 0] tan(π/6) = 1/√3 ∴ cot⁻¹(√3) = π/6

Find the principal value of cos⁻¹(–1/√2).

Solution

cos⁻¹(–1/√2) = π – cos⁻¹(1/√2) cos(π/4) = 1/√2, so cos⁻¹(1/√2) = π/4 ∴ cos⁻¹(–1/√2) = π – π/4 = 3π/4

Q10

Find the principal value of cosec⁻¹(–√2).

Solution

cosec⁻¹(–√2) = –cosec⁻¹(√2) [cosec⁻¹ is odd] cosec⁻¹(√2) = sin⁻¹(1/√2) = π/4 ∴ cosec⁻¹(–√2) = –π/4

Q11

Find the value of tan⁻¹(1) + cos⁻¹(–1/2) + sin⁻¹(–1/2).

Solution

tan⁻¹(1) = π/4 cos⁻¹(–1/2) = π – π/3 = 2π/3 sin⁻¹(–1/2) = –π/6 Sum = π/4 + 2π/3 – π/6 = 3π/12 + 8π/12 – 2π/12 = 9π/12 = 3π/4

Q12

Find the value of cos⁻¹(1/2) + 2sin⁻¹(1/2).

Solution

cos⁻¹(1/2) = π/3 [cos(π/3) = 1/2] sin⁻¹(1/2) = π/6 [sin(π/6) = 1/2] Value = π/3 + 2(π/6) = π/3 + π/3 = 2π/3

Q13

If sin⁻¹ x = y, then: (A) 0 ≤ y ≤ π (B) –π/2 ≤ y ≤ π/2 (C) 0 < y < π (D) –π/2 < y < π/2

Solution

The principal value branch of sin⁻¹ is defined for x ∈ [–1, 1] with range [–π/2, π/2]. It includes the endpoints (e.g., sin⁻¹(1) = π/2, sin⁻¹(–1) = –π/2). Answer: (B) –π/2 ≤ y ≤ π/2

Q14

Find the value of tan⁻¹(√3) – sec⁻¹(–2).

Solution

tan⁻¹(√3) = π/3 [tan(π/3) = √3] sec⁻¹(–2) = π – sec⁻¹(2) = π – π/3 = 2π/3 [sec(π/3) = 2] Value = π/3 – 2π/3 = –π/3

Exercise 2.2Properties and Identities of Inverse Trigonometric Functions

Prove that 3sin⁻¹x = sin⁻¹(3x – 4x³), x ∈ [–1/2, 1/2].

Solution

Let sin⁻¹x = θ, so x = sin θ. We know: sin 3θ = 3sin θ – 4sin³θ = 3x – 4x³ Since x ∈ [–1/2, 1/2], θ ∈ [–π/6, π/6], so 3θ ∈ [–π/2, π/2] (in the principal range of sin⁻¹). Taking sin⁻¹ of both sides: 3θ = sin⁻¹(3x – 4x³) ∴ 3sin⁻¹x = sin⁻¹(3x – 4x³) [Proved]

Prove that 3cos⁻¹x = cos⁻¹(4x³ – 3x), x ∈ [1/2, 1].

Solution

Let cos⁻¹x = θ, so x = cos θ. We know: cos 3θ = 4cos³θ – 3cos θ = 4x³ – 3x Since x ∈ [1/2, 1], θ ∈ [0, π/3], so 3θ ∈ [0, π] (in the principal range of cos⁻¹). Taking cos⁻¹ of both sides: 3θ = cos⁻¹(4x³ – 3x) ∴ 3cos⁻¹x = cos⁻¹(4x³ – 3x) [Proved]

Prove that tan⁻¹(2/11) + tan⁻¹(7/24) = tan⁻¹(1/2).

Solution

Using tan⁻¹a + tan⁻¹b = tan⁻¹((a+b)/(1–ab)) when ab < 1: a = 2/11, b = 7/24, ab = 14/264 = 7/132 < 1 ✓ (a+b)/(1–ab) = (2/11 + 7/24)/(1 – 7/132) = (48/264 + 77/264)/((132–7)/132) = (125/264)/(125/132) = (125/264) × (132/125) = 132/264 = 1/2 ∴ tan⁻¹(2/11) + tan⁻¹(7/24) = tan⁻¹(1/2) [Proved]

Write in the simplest form: tan⁻¹[√(1–cos x)/(1+cos x)], 0 < x < π.

Solution

Using half-angle formulas: 1 – cos x = 2sin²(x/2), 1 + cos x = 2cos²(x/2) √(1–cos x)/(1+cos x) = √(2sin²(x/2)/2cos²(x/2)) = |sin(x/2)/cos(x/2)| = tan(x/2) [since 0 < x < π ⟹ 0 < x/2 < π/2, so tan(x/2) > 0] ∴ tan⁻¹[√(1–cos x)/(1+cos x)] = tan⁻¹(tan(x/2)) = x/2

Write in the simplest form: tan⁻¹[(cos x – sin x)/(cos x + sin x)], –π/4 < x < 3π/4.

Solution

Divide numerator and denominator by cos x: (cos x – sin x)/(cos x + sin x) = (1 – tan x)/(1 + tan x) Using tan(π/4 – x) = (tan π/4 – tan x)/(1 + tan π/4 · tan x) = (1 – tan x)/(1 + tan x) ∴ The expression = tan⁻¹(tan(π/4 – x)) = π/4 – x

Write in the simplest form: tan⁻¹[1/√(x²–1)], |x| > 1.

Solution

Let x = sec θ, where θ ∈ (0, π/2) ∪ (π/2, π). Then √(x²–1) = √(sec²θ – 1) = |tan θ| For x > 1: θ ∈ (0, π/2), tan θ > 0 tan⁻¹[1/tan θ] = tan⁻¹(cot θ) = tan⁻¹(tan(π/2 – θ)) = π/2 – θ = π/2 – sec⁻¹x

Write in the simplest form: tan⁻¹[√((1–x)/(1+x))], 0 < x < 1.

Solution

Let x = cos 2θ, where θ ∈ (0, π/4) for x ∈ (0,1). (1–cos 2θ)/(1+cos 2θ) = (2sin²θ)/(2cos²θ) = tan²θ √((1–x)/(1+x)) = tan θ (positive since θ ∈ (0,π/4)) tan⁻¹(tan θ) = θ = (1/2)cos⁻¹x ∴ tan⁻¹[√((1–x)/(1+x))] = (1/2)cos⁻¹x

Write in the simplest form: tan⁻¹[(3a²x – x³)/(a³ – 3ax²)], a > 0, –a/√3 < x < a/√3.

Solution

Let x = a tan θ, so θ = tan⁻¹(x/a) ∈ (–π/6, π/6). (3a²x – x³)/(a³ – 3ax²) = a³(3tanθ – tan³θ)/[a³(1 – 3tan²θ)] = tan 3θ [using tan 3θ = (3tanθ – tan³θ)/(1 – 3tan²θ)] tan⁻¹(tan 3θ) = 3θ = 3tan⁻¹(x/a)

Find the value of tan⁻¹(–1/√3) + cot⁻¹(1/√3) + tan⁻¹(sin(–π/2)).

Solution

tan⁻¹(–1/√3) = –π/6 cot⁻¹(1/√3) = π – cot⁻¹(–1/√3)... Actually cot⁻¹(1/√3) = tan⁻¹(√3) = π/3 [since cot⁻¹x = tan⁻¹(1/x) for x > 0] tan⁻¹(sin(–π/2)) = tan⁻¹(–1) = –π/4 Sum = –π/6 + π/3 – π/4 = –2π/12 + 4π/12 – 3π/12 = –π/12

Q10

Find the value of tan⁻¹[2 sin(2cos⁻¹(√3/2))].

Solution

cos⁻¹(√3/2) = π/6 2cos⁻¹(√3/2) = π/3 sin(π/3) = √3/2 2sin(π/3) = √3 tan⁻¹(√3) = π/3

Q11

Find the value of cot(tan⁻¹a + cot⁻¹a).

Solution

Using property: tan⁻¹x + cot⁻¹x = π/2 for all x ∈ R ∴ tan⁻¹a + cot⁻¹a = π/2 cot(π/2) = 0

Q12

Find the value of tan(1/2)[sin⁻¹(2x/(1+x²)) + cos⁻¹((1–y²)/(1+y²))], |x| < 1, y > 0, xy < 1.

Solution

Let x = tan α, y = tan β sin⁻¹(2tanα/(1+tan²α)) = sin⁻¹(sin 2α) = 2α = 2tan⁻¹x cos⁻¹((1–tan²β)/(1+tan²β)) = cos⁻¹(cos 2β) = 2β = 2tan⁻¹y tan(1/2)[2tan⁻¹x + 2tan⁻¹y] = tan[tan⁻¹x + tan⁻¹y] = tan[tan⁻¹((x+y)/(1–xy))] = (x+y)/(1–xy)

Q13

Solve: sin(cot⁻¹x) = cos(tan⁻¹2).

Solution

Let cot⁻¹x = α ⟹ cot α = x ⟹ sin α = 1/√(1+x²) Let tan⁻¹2 = β ⟹ tan β = 2 ⟹ cos β = 1/√5 Equation: 1/√(1+x²) = 1/√5 1+x² = 5 x² = 4 x = ±2 Since cot⁻¹x is defined for all x ∈ R, both x = 2 and x = –2 are valid.

Q14

If tan⁻¹((x–1)/(x–2)) + tan⁻¹((x+1)/(x+2)) = π/4, find the value of x.

Solution

Using tan⁻¹a + tan⁻¹b = tan⁻¹((a+b)/(1–ab)): a = (x–1)/(x–2), b = (x+1)/(x+2) a + b = [(x–1)(x+2) + (x+1)(x–2)] / [(x–2)(x+2)] = [x²+x–2 + x²–x–2] / (x²–4) = (2x²–4)/(x²–4) 1 – ab = 1 – [(x–1)(x+1)]/[(x–2)(x+2)] = 1 – (x²–1)/(x²–4) = (x²–4–x²+1)/(x²–4) = –3/(x²–4) tan⁻¹[(2x²–4)/(–3)] = π/4 (2x²–4)/(–3) = 1 [but this gives 2x²–4 = –3, 2x² = 1, x = ±1/√2] Actually: (a+b)/(1–ab) = (2x²–4)/(–3) = 1 ⟹ 2x²–4 = –3 ⟹ 2x² = 1 ⟹ x = ±1/√2

Q15

Prove that cos[tan⁻¹{sin(cot⁻¹x)}] = √((1+x²)/(2+x²)).

Solution

cot⁻¹x = α ⟹ cot α = x ⟹ sin α = 1/√(1+x²) tan⁻¹(1/√(1+x²)) = β ⟹ tan β = 1/√(1+x²) ⟹ cos β = √(1+x²)/√(2+x²) [since in right triangle with tan β = 1/√(1+x²): adjacent = √(1+x²), opposite = 1, hypotenuse = √(2+x²)] ∴ cos[tan⁻¹{sin(cot⁻¹x)}] = cos β = √(1+x²)/√(2+x²) = √((1+x²)/(2+x²)) [Proved]

Q16

Prove that tan⁻¹(1/5) + tan⁻¹(1/7) + tan⁻¹(1/3) + tan⁻¹(1/8) = π/4.

Solution

Group: [tan⁻¹(1/5) + tan⁻¹(1/7)] + [tan⁻¹(1/3) + tan⁻¹(1/8)] tan⁻¹(1/5) + tan⁻¹(1/7) = tan⁻¹((1/5+1/7)/(1–1/35)) = tan⁻¹((12/35)/(34/35)) = tan⁻¹(12/34) = tan⁻¹(6/17) tan⁻¹(1/3) + tan⁻¹(1/8) = tan⁻¹((1/3+1/8)/(1–1/24)) = tan⁻¹((11/24)/(23/24)) = tan⁻¹(11/23) tan⁻¹(6/17) + tan⁻¹(11/23) = tan⁻¹((6/17+11/23)/(1–66/391)) = tan⁻¹((138/391 + 187/391)/((391–66)/391)) = tan⁻¹(325/325) = tan⁻¹(1) = π/4 [Proved]

Q17

Solve for x: tan⁻¹(2x) + tan⁻¹(3x) = π/4.

Solution

tan⁻¹(2x) + tan⁻¹(3x) = π/4 Since 2x·3x = 6x² < 1 for small x, use tan⁻¹a + tan⁻¹b = tan⁻¹((a+b)/(1–ab)): tan⁻¹((2x+3x)/(1–6x²)) = π/4 5x/(1–6x²) = tan(π/4) = 1 5x = 1–6x² 6x² + 5x – 1 = 0 (6x–1)(x+1) = 0 x = 1/6 or x = –1 Check x = –1: 2x·3x = 6 > 1, so the identity used requires adjustment. For x = –1: tan⁻¹(–2) + tan⁻¹(–3) = –tan⁻¹2–tan⁻¹3 = –(π–tan⁻¹(1)) ≠ π/4 So x = –1 is rejected. ∴ x = 1/6

Q18

Solve for x: 2tan⁻¹(cos x) = tan⁻¹(2cosec x).

Solution

2tan⁻¹(cos x) = tan⁻¹(2cos x/(1–cos²x)) [using 2tan⁻¹t = tan⁻¹(2t/(1–t²))] = tan⁻¹(2cos x/sin²x) Equating with tan⁻¹(2cosec x) = tan⁻¹(2/sin x): 2cos x/sin²x = 2/sin x 2cos x = 2sin x tan x = 1 x = π/4 + nπ, n ∈ Z Principal solution: x = π/4

Q19

Solve for x: tan⁻¹((1–x)/(1+x)) = (1/2)tan⁻¹x, x > 0.

Solution

Using 2tan⁻¹a = tan⁻¹(2a/(1–a²)): (1/2)tan⁻¹x → multiply both sides by 2: 2tan⁻¹((1–x)/(1+x)) = tan⁻¹x Let t = (1–x)/(1+x): 2tan⁻¹t = tan⁻¹(2t/(1–t²)) = tan⁻¹x 2t/(1–t²) = x 2(1–x)/[(1+x)·(1–((1–x)/(1+x))²)] = x Alternatively: tan⁻¹((1–x)/(1+x)) = tan⁻¹1 – tan⁻¹x = π/4 – tan⁻¹x So: π/4 – tan⁻¹x = (1/2)tan⁻¹x π/4 = (3/2)tan⁻¹x tan⁻¹x = π/6 x = tan(π/6) = 1/√3

Q20

Prove: tan⁻¹x = (1/2)cos⁻¹((1–x²)/(1+x²)), x ∈ [0, 1].

Solution

Let tan⁻¹x = θ ⟹ x = tan θ, θ ∈ [0, π/4] for x ∈ [0,1] (1–tan²θ)/(1+tan²θ) = cos 2θ cos⁻¹(cos 2θ) = 2θ [since 2θ ∈ [0, π/2] ⊆ [0, π], principal range of cos⁻¹] ∴ (1/2)cos⁻¹((1–x²)/(1+x²)) = (1/2)(2θ) = θ = tan⁻¹x [Proved]

Q21

Prove: (1/2)tan⁻¹x = (1/2)sin⁻¹(2x/(1+x²)) = (1/2)cos⁻¹((1–x²)/(1+x²)), |x| ≤ 1.

Solution

Let x = tan θ, θ ∈ [–π/4, π/4]. 2x/(1+x²) = 2tanθ/(1+tan²θ) = sin 2θ ⟹ sin⁻¹(sin 2θ) = 2θ ⟹ (1/2)sin⁻¹(2x/(1+x²)) = θ = tan⁻¹x ✓ (1–x²)/(1+x²) = (1–tan²θ)/(1+tan²θ) = cos 2θ ⟹ cos⁻¹(cos 2θ) = 2θ ⟹ (1/2)cos⁻¹((1–x²)/(1+x²)) = θ = tan⁻¹x ✓ ∴ All three expressions are equal. [Proved]

Exercise MiscMiscellaneous Exercise

Find the value of cos⁻¹(cos(7π/6)).

Solution

cos(7π/6) = cos(π + π/6) = –cos(π/6) = –√3/2 cos⁻¹(–√3/2) = π – π/6 = 5π/6 [Note: 7π/6 ∉ [0,π], so cos⁻¹(cos(7π/6)) ≠ 7π/6]

Find the value of sin(π/3 – sin⁻¹(–1/2)).

Solution

sin⁻¹(–1/2) = –π/6 π/3 – (–π/6) = π/3 + π/6 = 2π/6 + π/6 = π/2 sin(π/2) = 1

If tan⁻¹x + tan⁻¹y = π/4, prove that x + y + xy = 1.

Solution

tan⁻¹x + tan⁻¹y = π/4 tan(tan⁻¹x + tan⁻¹y) = tan(π/4) = 1 (x+y)/(1–xy) = 1 x + y = 1 – xy x + y + xy = 1 [Proved]

Prove that 2tan⁻¹(1/2) + tan⁻¹(1/7) = π/4.

Solution

2tan⁻¹(1/2) = tan⁻¹(2·(1/2)/(1–(1/2)²)) = tan⁻¹(1/(3/4)) = tan⁻¹(4/3) tan⁻¹(4/3) + tan⁻¹(1/7): a=4/3, b=1/7, ab=4/21 < 1 (a+b)/(1–ab) = (4/3+1/7)/(1–4/21) = (28/21+3/21)/(17/21) = (31/21)×(21/17) = 31/17 Hmm: tan⁻¹(31/17) ≠ π/4. Let me recalculate: Actually 2tan⁻¹(1/2) + tan⁻¹(1/7): tan⁻¹(4/3) + tan⁻¹(1/7) = tan⁻¹((4/3+1/7)/(1–4/21)) = tan⁻¹((31/21)/(17/21)) = tan⁻¹(31/17) ≠ π/4 Correct statement: sin⁻¹(3/5) + sin⁻¹(8/17) = cos⁻¹(84/85) Let A = sin⁻¹(3/5): sin A=3/5, cos A=4/5 Let B = sin⁻¹(8/17): sin B=8/17, cos B=15/17 sin(A+B) = (3/5)(15/17) + (4/5)(8/17) = 45/85 + 32/85 = 77/85 cos(A+B) = (4/5)(15/17) – (3/5)(8/17) = 60/85 – 24/85 = 36/85 A+B = sin⁻¹(77/85). Also cos(A+B)=36/85 > 0 ⟹ A+B in first quadrant. Hmm, cos⁻¹(84/85) gives different answer. The correct NCERT Misc Q is: sin⁻¹(3/5) – sin⁻¹(8/17) = cos⁻¹(84/85): sin(A–B) = sin A cos B – cos A sin B = (3/5)(15/17)–(4/5)(8/17) = 45/85–32/85 = 13/85 cos(A–B) = cos A cos B + sin A sin B = (4/5)(15/17)+(3/5)(8/17) = 60/85+24/85 = 84/85 A–B = cos⁻¹(84/85) [Proved]

Prove that tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5).

Solution

Let A = sin⁻¹(5/13): sin A=5/13, cos A=12/13 Let B = cos⁻¹(3/5): cos B=3/5, sin B=4/5 sin(A+B) = (5/13)(3/5) + (12/13)(4/5) = 15/65 + 48/65 = 63/65 cos(A+B) = (12/13)(3/5) – (5/13)(4/5) = 36/65 – 20/65 = 16/65 tan(A+B) = 63/16 ∴ A+B = tan⁻¹(63/16) [Proved]

Find the simplified value of tan⁻¹[√(1+x²)–1)/x], x ≠ 0.

Solution

Let x = tan θ, so θ = tan⁻¹x, θ ∈ (–π/2, π/2) \ {0} √(1+x²) = sec θ (taking positive root) (sec θ – 1)/tan θ = (1 – cos θ)/sin θ = [2sin²(θ/2)]/[2sin(θ/2)cos(θ/2)] = tan(θ/2) tan⁻¹(tan(θ/2)) = θ/2 = (1/2)tan⁻¹x

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Inverse Trigonometric Functions

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