MP Board Class 12 Maths Chapter 2: Inverse Trigonometric…

📘 MP Board Class 12 Maths Chapter 2: Inverse Trigonometric Functions – Notes, Formulas & Practice Questions

MP Board Class 12 Maths Chapter 2: प्रतिलोम त्रिकोणमितीय फलन (Inverse Trigonometric Functions) is an important chapter carrying 4–6 marks in the board exam. This chapter introduces the concept of inverse trigonometric functions, their domains, ranges, properties, and identities. Questions typically include finding principal values, proving identities, and solving equations. A strong grasp of this chapter is essential because inverse trigonometric functions are used extensively in calculus (Chapter 5 — Continuity & Differentiability and Chapter 7 — Integrals).

📑 Table of Contents

Introduction & Definition
Domains and Ranges of Inverse Trigonometric Functions
Principal Value Branches
Properties of Inverse Trigonometric Functions
Important Formulas & Identities
Practice Questions with Answers
Previous Year Questions (2017–2026)

🔢 Introduction & Definition

Trigonometric functions (sin, cos, tan, cot, sec, cosec) are periodic and hence not one-to-one over their entire domains. However, by restricting their domains to suitable intervals where they become one-to-one, we can define their inverse functions.

If y = f(x) = sin x, then its inverse is x = sin⁻¹(y), read as “sin inverse y” or “arc sine of y”.

Key Point: sin⁻¹ x is NOT equal to (sin x)⁻¹. It denotes the angle whose sine is x. (sin x)⁻¹ = 1/sin x = cosec x.

📌 Notation: sin⁻¹ x ∈ [−π/2, π/2] gives the principal value. Similarly for other functions with their respective principal value branches.

📊 Domains and Ranges of Inverse Trigonometric Functions

The following table gives you the domain (input values) and range (output/principal value) for each inverse trigonometric function — memorise this table, it is the most frequently tested topic.

Function	Domain (x)	Range (Principal Value)
y = sin⁻¹ x	[−1, 1]	[−π/2, π/2]
y = cos⁻¹ x	[−1, 1]	[0, π]
y = tan⁻¹ x	R (all real numbers)	(−π/2, π/2)
y = cot⁻¹ x	R	(0, π)
y = sec⁻¹ x	(−∞, −1] ∪ [1, ∞)	[0, π] − {π/2}
y = cosec⁻¹ x	(−∞, −1] ∪ [1, ∞)	[−π/2, π/2] − {0}

🎯 Principal Value — The Most Important Concept

The principal value of an inverse trigonometric function is the value that lies in its principal value branch (the range given in the table above). For example:

sin⁻¹(1/2) = π/6 (principal value), NOT 5π/6 or 13π/6 etc.
cos⁻¹(−1/2) = 2π/3 (principal value), NOT −2π/3 or 4π/3
tan⁻¹(1) = π/4 (principal value)

💡 Remember: For sin⁻¹ and tan⁻¹, the principal value always lies in the 1st or 4th quadrant (−π/2 to π/2). For cos⁻¹ and cot⁻¹, it lies in the 1st or 2nd quadrant (0 to π).

🔹 Steps to Find Principal Value

Identify the inverse function (e.g., sin⁻¹, cos⁻¹, tan⁻¹).
Recall the angle θ in the principal value range whose trigonometric ratio equals the given value.
Ensure the angle lies within the principal value range.
Write the answer.

🔗 Properties of Inverse Trigonometric Functions

🅰️ Property Set A — Cancellation Properties

These properties “cancel” the function and its inverse:

Property	Condition
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)

🅱️ Property Set B — Negative Arguments

Property	Domain
sin⁻¹(−x) = −sin⁻¹ x	x ∈ [−1, 1]
cos⁻¹(−x) = π − cos⁻¹ x	x ∈ [−1, 1]
tan⁻¹(−x) = −tan⁻¹ x	x ∈ R
cot⁻¹(−x) = π − cot⁻¹ x	x ∈ R
sec⁻¹(−x) = π − sec⁻¹ x	\|x\| ≥ 1
cosec⁻¹(−x) = −cosec⁻¹ x	\|x\| ≥ 1

🅲 Property Set C — Complementary (Co-function) Properties

Property	Domain
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

📝 Important Formulas & Identities

🔸 Sum and Difference Formulas

Formula	Condition
tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y)/(1 − xy))	xy < 1
tan⁻¹ x + tan⁻¹ y = π + tan⁻¹((x + y)/(1 − xy))	x > 0, y > 0, xy > 1
tan⁻¹ x + tan⁻¹ y = −π + tan⁻¹((x + y)/(1 − xy))	x < 0, y < 0, xy > 1
tan⁻¹ x − tan⁻¹ y = tan⁻¹((x − y)/(1 + xy))	Always valid
2 tan⁻¹ x = sin⁻¹(2x/(1+x²))	x ∈ [−1, 1]
2 tan⁻¹ x = cos⁻¹((1−x²)/(1+x²))	x ≥ 0
2 tan⁻¹ x = tan⁻¹(2x/(1−x²))	x ∈ (−1, 1)

🔸 Conversion Formulas (expressed as inverse trigonometric functions)

Formula	Domain
sin⁻¹ x = cosec⁻¹(1/x)	x ∈ [−1, 1], x ≠ 0
cos⁻¹ x = sec⁻¹(1/x)	x ∈ [−1, 1], x ≠ 0
tan⁻¹ x = cot⁻¹(1/x)	x > 0
sin⁻¹ x = cos⁻¹(√(1−x²))	x ∈ [0, 1]
cos⁻¹ x = sin⁻¹(√(1−x²))	x ∈ [0, 1]
tan⁻¹ x = sin⁻¹(x/√(1+x²))	x ∈ R
tan⁻¹ x = cos⁻¹(1/√(1+x²))	x ≥ 0

✏️ Practice Questions with Answers

🟢 Level 1 — Finding Principal Values

Q1. Find the principal value of sin⁻¹(−1/2).

Answer: −π/6 (or −30°). Since sin(−π/6) = −1/2 and −π/6 ∈ [−π/2, π/2].

Q2. Find the principal value of cos⁻¹(√3/2).

Answer: π/6. Since cos(π/6) = √3/2 and π/6 ∈ [0, π].

Q3. Find the principal value of tan⁻¹(−1).

Answer: −π/4. Since tan(−π/4) = −1 and −π/4 ∈ (−π/2, π/2).

Q4. Find the principal value of cosec⁻¹(2).

Answer: π/6. Since cosec(π/6) = 2 and π/6 ∈ [−π/2, π/2] − {0}.

🟡 Level 2 — Using Properties

Q5. Prove that sin⁻¹(3/5) + sin⁻¹(8/17) = sin⁻¹(77/85).

Answer: Let sin⁻¹(3/5) = α, sin⁻¹(8/17) = β. Then sin α = 3/5, cos α = 4/5; sin β = 8/17, cos β = 15/17. Now sin(α+β) = sin α cos β + cos α sin β = (3/5)(15/17) + (4/5)(8/17) = 45/85 + 32/85 = 77/85. Since α+β ∈ (0, π/2), sin⁻¹(77/85) = α+β. Hence proved.

Q6. Simplify: tan⁻¹(1/√(x²−1)), |x| > 1.

Answer: Put x = sec θ or x = cosec θ. If x = sec θ, θ = sec⁻¹ x. Then tan⁻¹(1/√(x²−1)) = tan⁻¹(1/tan θ) = tan⁻¹(cot θ) = tan⁻¹(tan(π/2 − θ)) = π/2 − θ = π/2 − sec⁻¹ x.

Q7. Prove that tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = π.

Answer: tan⁻¹(1) = π/4. Now tan⁻¹(2) + tan⁻¹(3) = π + tan⁻¹((2+3)/(1−6)) = π + tan⁻¹(−1) = π + (−π/4) = 3π/4. Adding π/4 gives π. Hence proved.

🔴 Level 3 — Advanced Problems

Q8. Solve for x: tan⁻¹(2x) + tan⁻¹(3x) = π/4.

Answer: Using tan⁻¹ a + tan⁻¹ b = tan⁻¹((a+b)/(1−ab)), we get tan⁻¹((2x+3x)/(1−6x²)) = π/4. So (5x)/(1−6x²) = tan(π/4) = 1. Thus 5x = 1−6x² → 6x²+5x−1=0 → (6x−1)(x+1)=0. So x = 1/6 or x = −1. But x = −1 gives LHS negative, invalid. Hence x = 1/6.

Q9. Express sin⁻¹(x√(1−x) + √x·√(1−x²)) in simplest form for 0 < x < 1.

Answer: Let sin⁻¹ x = θ, sin⁻¹ √x = φ. Then sin(θ+φ) = sin θ cos φ + cos θ sin φ = x·√(1−x) + √(1−x²)·√x. Hence the expression equals sin⁻¹ x + sin⁻¹(√x).

📋 Previous Year Questions (MP Board 2017–2026)

PYQ 1 (2024, 2 marks): Find the principal value of cos⁻¹(−√3/2).

Answer: 5π/6. Since cos(5π/6) = −√3/2 and 5π/6 ∈ [0, π].

PYQ 2 (2023, 2 marks): Find the principal value of cot⁻¹(−1/√3).

Answer: 2π/3. Since cot(2π/3) = cos(2π/3)/sin(2π/3) = (−1/2)/(√3/2) = −1/√3, and 2π/3 ∈ (0, π).

PYQ 3 (2022, 4 marks): Prove that tan⁻¹(1/4) + tan⁻¹(2/9) = (1/2)cos⁻¹(3/5).

Answer: LHS: tan⁻¹(1/4) + tan⁻¹(2/9) = tan⁻¹((1/4+2/9)/(1−1/18)) = tan⁻¹((9+8)/36/(17/18)) = tan⁻¹((17/36)×(18/17)) = tan⁻¹(1/2). RHS: (1/2)cos⁻¹(3/5). Let cos⁻¹(3/5) = θ. Then cos θ = 3/5, so tan(θ/2) = √((1−cosθ)/(1+cosθ)) = √((1−3/5)/(1+3/5)) = √(2/8) = 1/2. So θ/2 = tan⁻¹(1/2). Hence both sides equal tan⁻¹(1/2). Proved.

PYQ 4 (2021, 2 marks): Write the principal value of sin⁻¹(sin(2π/3)).

Answer: π/3. Since sin(2π/3) = sin(π/3), and sin⁻¹(sin x) = x only when x ∈ [−π/2, π/2]. Here 2π/3 is not in this range, so sin⁻¹(sin(2π/3)) = sin⁻¹(sin(π−π/3)) = sin⁻¹(sin(π/3)) = π/3.

PYQ 5 (2020, 4 marks): Solve: tan⁻¹(1−x)/(1+x) = (1/2)tan⁻¹ x, x > 0.

Answer: Given tan⁻¹((1−x)/(1+x)) = (1/2)tan⁻¹ x. Put x = tan θ. Then LHS = tan⁻¹((1−tanθ)/(1+tanθ)) = tan⁻¹(tan(π/4−θ)) = π/4−θ. RHS = (1/2)θ. So π/4−θ = θ/2 → π/4 = 3θ/2 → θ = π/6. Hence x = tan(π/6) = 1/√3.

PYQ 6 (2019, 2 marks): Find the principal value of sec⁻¹(−2/√3).

Answer: 5π/6. Since sec(5π/6) = 1/cos(5π/6) = 1/(−√3/2) = −2/√3, and 5π/6 ∈ [0,π]−{π/2}.

📌 Quick Revision Tips

Memorise the domain & range table — questions about principal values are guaranteed every year.
Remember that sin⁻¹, tan⁻¹, cosec⁻¹ give values in Q1 or Q4 (right half).
cos⁻¹, cot⁻¹, sec⁻¹ give values in Q1 or Q2 (upper half).
For identities, the tan⁻¹ sum formula with conditions is the most frequently tested.
Practice substitution problems — put x = tan θ, sin θ, sec θ to simplify expressions.
Chapter 2 connects directly to Ch 5 (Continuity & Differentiability) and Ch 7 (Integrals) — mastering this chapter is essential for calculus.

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