Class 12 Mathematics Notes Chapter 2 (Inverse trigonometric functions) – Mathematics Part-I Book

Detailed Notes with MCQs of Chapter 2: Inverse Trigonometric Functions. This is a crucial chapter, not just for your board exams but also forms the basis for many concepts in calculus and is frequently tested in government entrance exams. Pay close attention to the definitions, domains, principal value branches, and properties.

Chapter 2: Inverse Trigonometric Functions - Detailed Notes

1. Introduction: Why Inverse Trigonometric Functions?

• Trigonometric functions (sin, cos, tan, etc.) take an angle as input and give a ratio as output (e.g., sin(π/6) = 1/2).
• Inverse trigonometric functions do the opposite: they take a ratio as input and give an angle as output (e.g., sin⁻¹(1/2) = π/6).
• Notation: The inverse sine function is denoted by sin⁻¹x (or arcsin x). Important: sin⁻¹x does not mean (sin x)⁻¹ = 1/sin x.
• The Problem: Trigonometric functions are periodic (e.g., sin x repeats every 2π). This means they are many-to-one functions (different angles can have the same sine value, like sin 0 = 0, sin π = 0, sin 2π = 0). Functions must be one-to-one and onto (bijective) to have an inverse.
• The Solution: We restrict the domain of the original trigonometric functions so that they become bijective within that restricted domain. The range of the inverse function corresponds to this restricted domain.

2. Definitions, Domains, and Principal Value Branches

To make trigonometric functions invertible, we restrict their domains. The range of the corresponding inverse function is called the Principal Value Branch. This is extremely important!

• Principal Value: When asked to find the value of an inverse trigonometric function (e.g., find sin⁻¹(1/2)), you MUST give the answer that lies within the Principal Value Branch. For sin⁻¹(1/2), the answer is π/6, not 5π/6 or 13π/6, etc.

3. Graphs of Inverse Trigonometric Functions

• The graph of y = f⁻¹(x) is the reflection of the graph of y = f(x) (with its restricted domain) about the line y = x.
• Understanding the shape helps visualize the domain and range. (Refer to NCERT textbook for graphs).

4. Properties of Inverse Trigonometric Functions

These properties are essential for simplifying expressions and solving equations.

(A) Self-Adjusting Properties:

• sin(sin⁻¹x) = x, if x ∈ [-1, 1]

• cos(cos⁻¹x) = x, if x ∈ [-1, 1]

• tan(tan⁻¹x) = x, if x ∈ R

• cot(cot⁻¹x) = x, if x ∈ R

• sec(sec⁻¹x) = x, if x ∈ R - (-1, 1)

• cosec(cosec⁻¹x) = x, if x ∈ R - (-1, 1)

• sin⁻¹(sin x) = x, if x ∈ [-π/2, π/2] (Caution! If x is outside this range, adjust it first)

• cos⁻¹(cos x) = x, if x ∈ [0, π] (Caution!)

• tan⁻¹(tan x) = x, if x ∈ (-π/2, π/2) (Caution!)

• cot⁻¹(cot x) = x, if x ∈ (0, π) (Caution!)

• sec⁻¹(sec x) = x, if x ∈ [0, π] - {π/2} (Caution!)

• cosec⁻¹(cosec x) = x, if x ∈ [-π/2, π/2] - {0} (Caution!)

  Example: sin⁻¹(sin(2π/3)). Here, 2π/3 is not in [-π/2, π/2]. We know sin(2π/3) = sin(π - π/3) = sin(π/3). So, sin⁻¹(sin(2π/3)) = sin⁻¹(sin(π/3)) = π/3.

(B) Negative Argument Properties:

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

(C) Reciprocal Properties:

• cosec⁻¹x = sin⁻¹(1/x), |x| ≥ 1
• sec⁻¹x = cos⁻¹(1/x), |x| ≥ 1
• cot⁻¹x = tan⁻¹(1/x), x > 0
• cot⁻¹x = π + tan⁻¹(1/x), x < 0

(D) Complementary Angle Properties:

• sin⁻¹x + cos⁻¹x = π/2, x ∈ [-1, 1]
• tan⁻¹x + cot⁻¹x = π/2, x ∈ R
• sec⁻¹x + cosec⁻¹x = π/2, |x| ≥ 1

(E) Sum and Difference Formulas (Mainly for tan⁻¹):

• tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy)), if xy < 1

• tan⁻¹x + tan⁻¹y = π + tan⁻¹((x+y)/(1-xy)), if xy > 1 and x > 0, y > 0

• tan⁻¹x + tan⁻¹y = -π + tan⁻¹((x+y)/(1-xy)), if xy > 1 and x < 0, y < 0

• tan⁻¹x - tan⁻¹y = tan⁻¹((x-y)/(1+xy)), if xy > -1

• 2tan⁻¹x = tan⁻¹(2x / (1-x²)), if |x| < 1

• 2tan⁻¹x = sin⁻¹(2x / (1+x²)), if |x| ≤ 1

• 2tan⁻¹x = cos⁻¹((1-x²) / (1+x²)), if x ≥ 0

(F) Conversion Between Inverse Functions:

You can convert one inverse trigonometric function to another using a right-angled triangle.
Example: Convert sin⁻¹(3/5) to tan⁻¹.
Let θ = sin⁻¹(3/5). Then sin θ = 3/5 (Opposite/Hypotenuse).
Using Pythagoras theorem, Adjacent = √(5² - 3²) = √16 = 4.
So, tan θ = Opposite/Adjacent = 3/4.
Therefore, sin⁻¹(3/5) = tan⁻¹(3/4). (Ensure the angle θ lies in the correct principal value branch for the target function).

5. Common Problem Types in Exams:

• Finding the Principal Value of expressions like cos⁻¹(-1/2), tan⁻¹(√3), sin⁻¹(sin(7π/6)).
• Proving identities using the properties.
• Simplifying complex expressions involving inverse trigonometric functions.
• Solving equations involving inverse trigonometric functions (e.g., solve tan⁻¹(2x) + tan⁻¹(3x) = π/4).

Key Takeaways for Government Exams:

• Memorize the Domain and Principal Value Branch table perfectly. Many direct questions come from this.
• Master the properties, especially negative arguments, complementary angles, and tan⁻¹ sum/difference formulas (including conditions).
• Be very careful with the condition for sin⁻¹(sin x) = x, cos⁻¹(cos x) = x etc. Always bring the angle within the principal value branch first.
• Practice converting between different inverse functions using the triangle method.

Multiple Choice Questions (MCQs)

1. The principal value of cos⁻¹(-√3/2) is:
   (a) -π/6
   (b) π/6
   (c) 5π/6
   (d) 7π/6

2. The value of sin⁻¹(sin(5π/4)) is:
   (a) 5π/4
   (b) π/4
   (c) -π/4
   (d) 3π/4

3. The domain of the function f(x) = sin⁻¹(2x - 3) is:
   (a) [-1, 1]
   (b) [1, 2]
   (c) [-2, -1]
   (d) [0, 1]

4. The value of tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2) is:
   (a) π/4
   (b) π/2
   (c) 3π/4
   (d) π

5. If sin⁻¹x + sin⁻¹y = π/2, then the value of cos⁻¹x + cos⁻¹y is:
   (a) π/2
   (b) π
   (c) 0
   (d) 2π/3

6. The value of tan(cos⁻¹(4/5) + tan⁻¹(2/3)) is:
   (a) 6/17
   (b) 7/16
   (c) 16/7
   (d) 17/6

7. The principal value branch of sec⁻¹x is:
   (a) [-π/2, π/2] - {0}
   (b) [0, π] - {π/2}
   (c) (0, π)
   (d) (-π/2, π/2)

8. The value of 2tan⁻¹(1/3) + tan⁻¹(1/7) is equal to:
   (a) tan⁻¹(1)
   (b) π/2
   (c) π/4
   (d) tan⁻¹(2)

9. cot⁻¹(-√3) equals:
   (a) -π/6
   (b) π/6
   (c) 2π/3
   (d) 5π/6

10. The value of cos(sec⁻¹x + cosec⁻¹x), where |x| ≥ 1, is:
    (a) 1
    (b) -1
    (c) 0
    (d) π/2

Answer Key:

1. (c) [Hint: cos⁻¹(-x) = π - cos⁻¹x. cos⁻¹(√3/2) = π/6. So, π - π/6 = 5π/6]
2. (c) [Hint: sin(5π/4) = sin(π + π/4) = -sin(π/4). sin⁻¹(-sin(π/4)) = -sin⁻¹(sin(π/4)) = -π/4]
3. (b) [Hint: Domain of sin⁻¹ is [-1, 1]. So, -1 ≤ 2x - 3 ≤ 1. Add 3: 2 ≤ 2x ≤ 4. Divide by 2: 1 ≤ x ≤ 2]
4. (c) [Hint: π/4 + (π - π/3) + (-π/6) = π/4 + 2π/3 - π/6 = (3π + 8π - 2π)/12 = 9π/12 = 3π/4]
5. (a) [Hint: sin⁻¹x = π/2 - cos⁻¹x and sin⁻¹y = π/2 - cos⁻¹y. Substitute into the given equation: (π/2 - cos⁻¹x) + (π/2 - cos⁻¹y) = π/2 => π - (cos⁻¹x + cos⁻¹y) = π/2 => cos⁻¹x + cos⁻¹y = π - π/2 = π/2]
6. (d) [Hint: Let α = cos⁻¹(4/5). Then cos α = 4/5, sin α = 3/5, tan α = 3/4. We need tan(α + tan⁻¹(2/3)). Let β = tan⁻¹(2/3). tan(α+β) = (tan α + tan β)/(1 - tan α tan β) = (3/4 + 2/3) / (1 - (3/4)(2/3)) = ((9+8)/12) / (1 - 6/12) = (17/12) / (6/12) = 17/6]
7. (b) [Direct definition]
8. (c) [Hint: Use 2tan⁻¹x = tan⁻¹(2x/(1-x²)). 2tan⁻¹(1/3) = tan⁻¹(2(1/3)/(1-(1/3)²)) = tan⁻¹((2/3)/(8/9)) = tan⁻¹(6/8) = tan⁻¹(3/4). Now, tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹((3/4 + 1/7)/(1 - (3/4)(1/7))) = tan⁻¹(((21+4)/28)/((28-3)/28)) = tan⁻¹(25/25) = tan⁻¹(1) = π/4]
9. (d) [Hint: cot⁻¹(-x) = π - cot⁻¹x. cot⁻¹(√3) = π/6. So, π - π/6 = 5π/6]
10. (c) [Hint: sec⁻¹x + cosec⁻¹x = π/2 for |x| ≥ 1. So, cos(π/2) = 0]

Study these notes thoroughly, understand the concepts behind the properties, and practice solving various types of problems. Good luck!