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Inverse Trigonometric Functions
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Class 12MathsChapter 2NCERTCBSE
Inverse Trigonometric Functions – Class 12 Maths Chapter 2
📌 Quick Overview: Chapter 2 deals with the inverse of trigonometric functions — sin⁻¹, cos⁻¹, tan⁻¹, cot⁻¹, sec⁻¹, cosec⁻¹. You will learn their domains, ranges (principal value branches), and important properties. This chapter is heavily tested in CBSE board exams — especially principal values and properties.
📋 Topics Covered:
- > Why Inverse Trig Functions?
- > sin⁻¹ (arcsin)
- > cos⁻¹ (arccos)
- > tan⁻¹ (arctan)
- > cot⁻¹ (arccot)
- > sec⁻¹ (arcsec)
- > cosec⁻¹ (arccosec)
- > Domain and Range Table
- > Principal Value Branch
- > Properties (all groups)
- > Solved Board Questions
- > MCQ Practice
1. Why Do We Need Inverse Trigonometric Functions?
Trigonometric functions like sin, cos, tan are not one-one on their natural domains (because they are periodic). To define their inverses, we restrict their domains so they become bijective. The resulting inverses are called inverse trigonometric functions or arc functions.
⚠️ Key Point: sin⁻¹(x) does NOT mean 1/sin(x). It means the angle whose sine is x. It is also written as arcsin(x).
2. Domain and Range (Principal Value Branch) – Master Table ⭐
| Function | Domain | Range (Principal Value Branch) |
|---|---|---|
| sin⁻¹(x) | [−1, 1] | [−π/2, π/2] |
| cos⁻¹(x) | [−1, 1] | [0, π] |
| tan⁻¹(x) | (−∞, ∞) = R | (−π/2, π/2) |
| cot⁻¹(x) | (−∞, ∞) = R | (0, π) |
| sec⁻¹(x) | (−∞, −1] ∪ [1, ∞) | [0, π] − {π/2} |
| cosec⁻¹(x) | (−∞, −1] ∪ [1, ∞) | [−π/2, π/2] − {0} |
⭐ Memory Trick:
• sin⁻¹ and tan⁻¹ → range is symmetric around 0: [−π/2, π/2]
• cos⁻¹ and cot⁻¹ → range starts from 0: [0, π] or (0, π)
• sec⁻¹ excludes π/2, cosec⁻¹ excludes 0.
• sin⁻¹ and tan⁻¹ → range is symmetric around 0: [−π/2, π/2]
• cos⁻¹ and cot⁻¹ → range starts from 0: [0, π] or (0, π)
• sec⁻¹ excludes π/2, cosec⁻¹ excludes 0.
3. Finding Principal Values
The principal value of an inverse trig function is the unique value in its principal value branch (range).
Example 1: Find sin⁻¹(1/2).
Let sin⁻¹(1/2) = θ ⟹ sin θ = 1/2 and θ ∈ [−π/2, π/2]
⟹ θ = π/6
Let sin⁻¹(1/2) = θ ⟹ sin θ = 1/2 and θ ∈ [−π/2, π/2]
⟹ θ = π/6
Example 2: Find cos⁻¹(−1/2).
Let cos⁻¹(−1/2) = θ ⟹ cos θ = −1/2 and θ ∈ [0, π]
cos(2π/3) = −1/2 ⟹ θ = 2π/3
Let cos⁻¹(−1/2) = θ ⟹ cos θ = −1/2 and θ ∈ [0, π]
cos(2π/3) = −1/2 ⟹ θ = 2π/3
Example 3: Find tan⁻¹(−1).
Let tan⁻¹(−1) = θ ⟹ tan θ = −1 and θ ∈ (−π/2, π/2)
⟹ θ = −π/4
Let tan⁻¹(−1) = θ ⟹ tan θ = −1 and θ ∈ (−π/2, π/2)
⟹ θ = −π/4
Example 4: Find cos⁻¹(cos 7π/6).
7π/6 ∉ [0, π], so we simplify: cos(7π/6) = cos(2π − 5π/6) = cos(5π/6)
Wait — cos(7π/6) = −cos(π/6) = −√3/2, and cos⁻¹(−√3/2) = 5π/6 ∈ [0, π]
⟹ Answer = 5π/6
7π/6 ∉ [0, π], so we simplify: cos(7π/6) = cos(2π − 5π/6) = cos(5π/6)
Wait — cos(7π/6) = −cos(π/6) = −√3/2, and cos⁻¹(−√3/2) = 5π/6 ∈ [0, π]
⟹ Answer = 5π/6
4. Standard Values – Quick Reference
| Value | sin⁻¹ | cos⁻¹ | tan⁻¹ |
|---|---|---|---|
| 0 | 0 | π/2 | 0 |
| 1/2 | π/6 | π/3 | — |
| 1/√2 | π/4 | π/4 | — |
| √3/2 | π/3 | π/6 | — |
| 1 | π/2 | 0 | π/4 |
| −1/2 | −π/6 | 2π/3 | — |
| −1/√2 | −π/4 | 3π/4 | — |
| −√3/2 | −π/3 | 5π/6 | — |
| −1 | −π/2 | π | −π/4 |
| 1/√3 | — | — | π/6 |
| √3 | — | — | π/3 |
5. Properties of Inverse Trigonometric Functions ⭐
Group 1: Reciprocal Relations
sin⁻¹(1/x) = cosec⁻¹(x), x ≥ 1 or x ≤ −1
cos⁻¹(1/x) = sec⁻¹(x), x ≥ 1 or x ≤ −1
tan⁻¹(1/x) = cot⁻¹(x), x > 0
tan⁻¹(1/x) = −π + cot⁻¹(x), x < 0
cos⁻¹(1/x) = sec⁻¹(x), x ≥ 1 or x ≤ −1
tan⁻¹(1/x) = cot⁻¹(x), x > 0
tan⁻¹(1/x) = −π + cot⁻¹(x), x < 0
Group 2: Negative Argument
sin⁻¹(−x) = −sin⁻¹(x), x ∈ [−1, 1]
tan⁻¹(−x) = −tan⁻¹(x), x ∈ R
cosec⁻¹(−x) = −cosec⁻¹(x), |x| ≥ 1
cos⁻¹(−x) = π − cos⁻¹(x), x ∈ [−1, 1]
sec⁻¹(−x) = π − sec⁻¹(x), |x| ≥ 1
cot⁻¹(−x) = π − cot⁻¹(x), x ∈ R
tan⁻¹(−x) = −tan⁻¹(x), x ∈ R
cosec⁻¹(−x) = −cosec⁻¹(x), |x| ≥ 1
cos⁻¹(−x) = π − cos⁻¹(x), x ∈ [−1, 1]
sec⁻¹(−x) = π − sec⁻¹(x), |x| ≥ 1
cot⁻¹(−x) = π − cot⁻¹(x), x ∈ R
⭐ Trick: sin⁻¹, tan⁻¹, cosec⁻¹ are odd functions → negative argument gives negative value.
cos⁻¹, sec⁻¹, cot⁻¹ → negative argument gives π minus the positive value.
cos⁻¹, sec⁻¹, cot⁻¹ → negative argument gives π minus the positive value.
Group 3: Complementary Relations
sin⁻¹(x) + cos⁻¹(x) = π/2, x ∈ [−1, 1]
tan⁻¹(x) + cot⁻¹(x) = π/2, x ∈ R
sec⁻¹(x) + cosec⁻¹(x) = π/2, |x| ≥ 1
tan⁻¹(x) + cot⁻¹(x) = π/2, x ∈ R
sec⁻¹(x) + cosec⁻¹(x) = π/2, |x| ≥ 1
Group 4: Composition with Trig Functions
sin(sin⁻¹(x)) = x, x ∈ [−1, 1]
cos(cos⁻¹(x)) = x, x ∈ [−1, 1]
tan(tan⁻¹(x)) = x, x ∈ R
sin⁻¹(sin x) = x, x ∈ [−π/2, π/2]
cos⁻¹(cos x) = x, x ∈ [0, π]
tan⁻¹(tan x) = x, x ∈ (−π/2, π/2)
cos(cos⁻¹(x)) = x, x ∈ [−1, 1]
tan(tan⁻¹(x)) = x, x ∈ R
sin⁻¹(sin x) = x, x ∈ [−π/2, π/2]
cos⁻¹(cos x) = x, x ∈ [0, π]
tan⁻¹(tan x) = x, x ∈ (−π/2, π/2)
⚠️ Warning: sin⁻¹(sin x) = x ONLY when x is in [−π/2, π/2]. If x is outside this range, simplify first!
Example: sin⁻¹(sin 2π/3) ≠ 2π/3. Since 2π/3 ∉ [−π/2, π/2], we write sin(2π/3) = sin(π/3) = √3/2, so sin⁻¹(sin 2π/3) = π/3.
Example: sin⁻¹(sin 2π/3) ≠ 2π/3. Since 2π/3 ∉ [−π/2, π/2], we write sin(2π/3) = sin(π/3) = √3/2, so sin⁻¹(sin 2π/3) = π/3.
Group 5: Addition Formulas ⭐ (Most Important for Board)
tan⁻¹(x) + tan⁻¹(y) = tan⁻¹[(x+y)/(1−xy)], if xy < 1
tan⁻¹(x) + tan⁻¹(y) = π + tan⁻¹[(x+y)/(1−xy)], if xy > 1 and x > 0, y > 0
tan⁻¹(x) − tan⁻¹(y) = tan⁻¹[(x−y)/(1+xy)], if xy > −1
2tan⁻¹(x) = sin⁻¹[2x/(1+x²)] = cos⁻¹[(1−x²)/(1+x²)] = tan⁻¹[2x/(1−x²)]
tan⁻¹(x) + tan⁻¹(y) = π + tan⁻¹[(x+y)/(1−xy)], if xy > 1 and x > 0, y > 0
tan⁻¹(x) − tan⁻¹(y) = tan⁻¹[(x−y)/(1+xy)], if xy > −1
2tan⁻¹(x) = sin⁻¹[2x/(1+x²)] = cos⁻¹[(1−x²)/(1+x²)] = tan⁻¹[2x/(1−x²)]
6. Important Board Exam Solved Questions
Q1. Find the principal value of sin⁻¹(−√3/2).
Let sin⁻¹(−√3/2) = θ ⟹ sin θ = −√3/2, θ ∈ [−π/2, π/2]
We know sin(π/3) = √3/2, so sin(−π/3) = −√3/2
∴ θ = −π/3
We know sin(π/3) = √3/2, so sin(−π/3) = −√3/2
∴ θ = −π/3
Q2. Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
tan⁻¹(1) = π/4
cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3
sin⁻¹(−1/2) = −sin⁻¹(1/2) = −π/6
Total = π/4 + 2π/3 − π/6 = 3π/12 + 8π/12 − 2π/12 = 9π/12 = 3π/4
cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3
sin⁻¹(−1/2) = −sin⁻¹(1/2) = −π/6
Total = π/4 + 2π/3 − π/6 = 3π/12 + 8π/12 − 2π/12 = 9π/12 = 3π/4
Q3. Prove that: tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4).
LHS = tan⁻¹[(1/2 + 2/11)/(1 − 1/2 × 2/11)]
= tan⁻¹[(11/22 + 4/22)/(1 − 2/22)]
= tan⁻¹[(15/22)/(20/22)]
= tan⁻¹[15/20]
= tan⁻¹(3/4) = RHS ✓
= tan⁻¹[(11/22 + 4/22)/(1 − 2/22)]
= tan⁻¹[(15/22)/(20/22)]
= tan⁻¹[15/20]
= tan⁻¹(3/4) = RHS ✓
Q4. Find the value of sin[π/3 − sin⁻¹(−1/2)].
sin⁻¹(−1/2) = −π/6
∴ sin[π/3 − (−π/6)] = sin[π/3 + π/6] = sin[2π/6 + π/6] = sin[3π/6] = sin(π/2) = 1
∴ sin[π/3 − (−π/6)] = sin[π/3 + π/6] = sin[2π/6 + π/6] = sin[3π/6] = sin(π/2) = 1
Q5. Solve for x: tan⁻¹(2x) + tan⁻¹(3x) = π/4.
Using tan⁻¹(a) + tan⁻¹(b) = tan⁻¹[(a+b)/(1−ab)] when ab < 1:
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 makes product 2×3×(−1)² = 6 > 1, so we use the π formula — reject for π/4 case.
∴ x = 1/6
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 makes product 2×3×(−1)² = 6 > 1, so we use the π formula — reject for π/4 case.
∴ x = 1/6
Q6. Write in simplest form: tan⁻¹[√(1−cos x)/√(1+cos x)].
Use identities: 1 − cos x = 2sin²(x/2), 1 + cos x = 2cos²(x/2)
= tan⁻¹[√(2sin²(x/2)) / √(2cos²(x/2))]
= tan⁻¹[sin(x/2)/cos(x/2)]
= tan⁻¹[tan(x/2)]
= x/2 (when x ∈ (0, π))
= tan⁻¹[√(2sin²(x/2)) / √(2cos²(x/2))]
= tan⁻¹[sin(x/2)/cos(x/2)]
= tan⁻¹[tan(x/2)]
= x/2 (when x ∈ (0, π))
7. All Key Formulas at a Glance
📌 DOMAIN & RANGE:
sin⁻¹: Domain [−1,1], Range [−π/2, π/2]
cos⁻¹: Domain [−1,1], Range [0, π]
tan⁻¹: Domain R, Range (−π/2, π/2)
📌 COMPLEMENTARY:
sin⁻¹x + cos⁻¹x = π/2
tan⁻¹x + cot⁻¹x = π/2
sec⁻¹x + cosec⁻¹x = π/2
📌 NEGATIVE:
sin⁻¹(−x) = −sin⁻¹(x)
cos⁻¹(−x) = π − cos⁻¹(x)
tan⁻¹(−x) = −tan⁻¹(x)
📌 ADDITION:
tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)], xy < 1
📌 DOUBLE:
2tan⁻¹x = sin⁻¹[2x/(1+x²)] = cos⁻¹[(1−x²)/(1+x²)]
8. MCQ Practice (1 Mark)
1. The principal value of cos⁻¹(√3/2) is:
(a) π/3 (b) π/6 (c) 5π/6 (d) π/4
Answer: (b) π/6
2. The domain of sin⁻¹(2x) is:
(a) [−1, 1] (b) [−1/2, 1/2] (c) [−2, 2] (d) R
Answer: (b) [−1/2, 1/2]
3. sin⁻¹(sin 3π/5) equals:
(a) 3π/5 (b) 2π/5 (c) π/5 (d) −3π/5
Answer: (b) 2π/5 [since sin(3π/5) = sin(π − 3π/5) = sin(2π/5)]
4. The value of tan⁻¹(√3) − cot⁻¹(−√3) is:
(a) π (b) −π/2 (c) 0 (d) 2√3
Answer: (b) −π/2 [tan⁻¹(√3) = π/3; cot⁻¹(−√3) = π − cot⁻¹(√3) = π − π/6 = 5π/6; result = π/3 − 5π/6 = −π/2]
9. Exam Tips 📝
- > Always check whether x is in the principal value range before writing the answer directly.
- > For sin⁻¹(sin θ), cos⁻¹(cos θ) — if angle is outside the principal range, convert using trig identities first.
- > Memorise the complementary pair: sin⁻¹x + cos⁻¹x = π/2 — it saves a lot of steps in questions.
- > In addition formulas, always check if xy < 1 or xy > 1 before applying the formula.
- > "Simplest form" questions almost always use trig half-angle substitutions — practice these well.
- > For "prove that" questions, always start from LHS and simplify step by step.
✅ Summary: Inverse Trig Functions is a scoring chapter if you memorise the domain-range table, 3 complementary pairs, and the 3 negative-argument rules. Practice principal value questions and addition formula problems — these are the most common board exam types. You've got this! 💪
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