Class 12 Mathematics Notes Chapter 2 (Inverse Trignometric Function) – Examplar Problems (English) Book

Detailed Notes with MCQs of Chapter 2: Inverse Trigonometric Functions. This is a crucial topic, not just for your board exams but also frequently tested in various government recruitment exams where mathematics is a component. Pay close attention to the definitions, domains, ranges (especially the Principal Value Branches), and properties.

Inverse Trigonometric Functions: Detailed Notes

1. Introduction & Basic Concept

• Why Inverse? Trigonometric functions (sin, cos, tan, etc.) are periodic, meaning they repeat their values. For a function to have an inverse, it must be bijective (both one-one and onto). Standard trigonometric functions are not one-one over their entire domain.
• Restriction: To make them invertible, we restrict their domains to intervals where they are one-one and onto. The range of the original function (in the restricted domain) becomes the domain of the inverse function, and the restricted domain of the original function becomes the range (called the Principal Value Branch) of the inverse function.
• Notation: The inverse of sin x is denoted by sin⁻¹x (or arcsin x), not (sin x)⁻¹ which is 1/sin x = cosec x. Similarly for cos⁻¹x (arccos x), tan⁻¹x (arctan x), etc.
• Meaning: If f(x) = y, then f⁻¹(y) = x. So, if sin θ = x, then sin⁻¹x = θ. Essentially, sin⁻¹x asks the question: "What angle (within the principal value branch) has a sine equal to x?"

2. Principal Value Branches (PVB)

This is the most critical part to memorize accurately. The value of an inverse trigonometric function calculated using calculators or standard tables lies within this range.

• Key Points about PVB:
  • sin⁻¹, tan⁻¹, cosec⁻¹ have ranges centered around 0 in Quadrants I and IV.
  • cos⁻¹, cot⁻¹, sec⁻¹ have ranges in Quadrants I and II.
  • Note the open vs. closed intervals and exclusions carefully.

3. Graphs of Inverse Trigonometric Functions

• The graph of y = f⁻¹(x) is the reflection of the graph of y = f(x) (in its restricted domain) across the line y = x.
• Understanding the shape helps visualize the range (PVB).

4. Fundamental Properties & Identities

(Assume x is within the valid domain for each function)

• Property 1 (Inverse of Function):

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

• Property 2 (Function of Inverse):

  • sin⁻¹(sin θ) = θ, only if θ ∈ [-π/2, π/2]
  • cos⁻¹(cos θ) = θ, only if θ ∈ [0, π]
  • tan⁻¹(tan θ) = θ, only if θ ∈ (-π/2, π/2)
  • cosec⁻¹(cosec θ) = θ, only if θ ∈ [-π/2, π/2] - {0}
  • sec⁻¹(sec θ) = θ, only if θ ∈ [0, π] - {π/2}
  • cot⁻¹(cot θ) = θ, only if θ ∈ (0, π)
  • Caution: If θ is outside the PVB, you must first use trigonometric identities to bring the angle within the PVB before applying the property. E.g., sin⁻¹(sin(2π/3)) = sin⁻¹(sin(π - π/3)) = sin⁻¹(sin(π/3)) = π/3.

• Property 3 (Negative Arguments):

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

• Property 4 (Reciprocal Relations):

  • sin⁻¹(1/x) = cosec⁻¹x, for x ∈ R - (-1, 1)
  • cos⁻¹(1/x) = sec⁻¹x, for x ∈ R - (-1, 1)
  • tan⁻¹(1/x) = cot⁻¹x, for x > 0
  • tan⁻¹(1/x) = -π + cot⁻¹x, for x < 0

• Property 5 (Complementary Angles):

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

5. Sum and Difference Formulas

• Most Important:
  • 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
• Other Useful Formulas:
  • 2tan⁻¹x = sin⁻¹(2x / (1+x²)), for |x| ≤ 1
  • 2tan⁻¹x = cos⁻¹((1-x²) / (1+x²)), for x ≥ 0
  • 2tan⁻¹x = tan⁻¹(2x / (1-x²)), for -1 < x < 1
• Formulas for sin⁻¹x ± sin⁻¹y and cos⁻¹x ± cos⁻¹y exist but are often more complex. It's frequently easier to convert to tan⁻¹ first, if possible.
  • sin⁻¹x + sin⁻¹y = sin⁻¹(x√(1-y²) + y√(1-x²))
  • sin⁻¹x - sin⁻¹y = sin⁻¹(x√(1-y²) - y√(1-x²))
  • cos⁻¹x + cos⁻¹y = cos⁻¹(xy - √(1-x²)√(1-y²))
  • cos⁻¹x - cos⁻¹y = cos⁻¹(xy + √(1-x²)√(1-y²))
  • Caution: These have conditions on x, y, and x²+y² for the results to lie in the PVB. Always verify or convert to tan⁻¹.

6. Conversion Between Inverse Functions

• This is a very useful technique for simplification and evaluation.
• Use a right-angled triangle. For example, to convert sin⁻¹(3/5) to other ITFs:
  1. Let θ = sin⁻¹(3/5). Then sin θ = 3/5 (Opposite/Hypotenuse).
  2. Draw a right triangle with opposite side 3, hypotenuse 5.
  3. By Pythagoras theorem, adjacent side = √(5² - 3²) = √16 = 4.
  4. Now, cos θ = Adj/Hyp = 4/5 => θ = cos⁻¹(4/5)
  5. tan θ = Opp/Adj = 3/4 => θ = tan⁻¹(3/4)
  6. So, sin⁻¹(3/5) = cos⁻¹(4/5) = tan⁻¹(3/4).

7. Solving Equations Involving ITFs

1. Simplify using properties.
2. Apply trigonometric functions to both sides (carefully).
3. Solve the resulting algebraic equation.
4. Crucially: Verify if the solutions obtained lie within the domain of the original ITF equation. Extraneous solutions are common.

Exam Preparation Tips:

• Master the Principal Value Branches. Many direct questions rely on this.
• Memorize Properties 3, 4, and 5 thoroughly.
• Practice the tan⁻¹x ± tan⁻¹y and 2tan⁻¹x formulas extensively.
• Be comfortable with the triangle method for conversions.
• Always check domain/range restrictions and conditions for formulas.
• When simplifying expressions like sin⁻¹(sin θ), always check if θ is in the PVB first.

Multiple Choice Questions (MCQs)

1. The principal value of cos⁻¹(-1/2) is:
   (A) -π/3
   (B) π/3
   (C) 2π/3
   (D) -2π/3

2. The value of sin⁻¹(sin(7π/6)) is:
   (A) 7π/6
   (B) π/6
   (C) -π/6
   (D) 5π/6

3. If sin⁻¹x + cos⁻¹(1/2) = π/2, then x is equal to:
   (A) 1/2
   (B) √3/2
   (C) -1/2
   (D) 1

4. The value of tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2) is:
   (A) π/4
   (B) 3π/4
   (C) π/2
   (D) π

5. 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

6. The domain of the function f(x) = sin⁻¹(2x - 3) is:
   (A) [-1, 1]
   (B) [1, 2]
   (C) [-1/2, 1/2]
   (D) [0, 1]

7. 2tan⁻¹(1/3) is equal to:
   (A) tan⁻¹(3/4)
   (B) tan⁻¹(2/3)
   (C) tan⁻¹(3/5)
   (D) sin⁻¹(3/5)

8. The value of cot(sin⁻¹x) is:
   (A) √(1-x²)/x
   (B) x/√(1-x²)
   (C) 1/x
   (D) √(1+x²)/x

9. If tan⁻¹(2) + tan⁻¹(3) = θ, then tan θ is:
   (A) -1
   (B) 1
   (C) 5
   (D) Undefined (Note: Be careful with formula conditions)

10. The value of sec⁻¹(-2/√3) is:
    (A) -π/6
    (B) π/6
    (C) 5π/6
    (D) 7π/6

Answers to MCQs:

1. (C) cos⁻¹(-x) = π - cos⁻¹x. So, cos⁻¹(-1/2) = π - cos⁻¹(1/2) = π - π/3 = 2π/3.
2. (C) 7π/6 is not in [-π/2, π/2]. sin(7π/6) = sin(π + π/6) = -sin(π/6) = -1/2. sin⁻¹(-1/2) = -sin⁻¹(1/2) = -π/6.
3. (A) We know sin⁻¹x + cos⁻¹x = π/2. Comparing with the given equation, cos⁻¹(1/2) must be equal to cos⁻¹x. So, x = 1/2.
4. (B) tan⁻¹(1) = π/4. cos⁻¹(-1/2) = 2π/3. sin⁻¹(-1/2) = -π/6. Sum = π/4 + 2π/3 - π/6 = (3π + 8π - 2π)/12 = 9π/12 = 3π/4.
5. (D) Let α = cos⁻¹(4/5). Then cos α = 4/5. Using a triangle (Adj=4, Hyp=5), Opp = 3. So tan α = 3/4. Let β = tan⁻¹(2/3). We need 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.
6. (B) For sin⁻¹(y), the domain is -1 ≤ y ≤ 1. So, -1 ≤ 2x - 3 ≤ 1. Add 3: 2 ≤ 2x ≤ 4. Divide by 2: 1 ≤ x ≤ 2. Domain is [1, 2].
7. (A) 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) / (1 - 1/9)) = tan⁻¹((2/3) / (8/9)) = tan⁻¹((2/3) * (9/8)) = tan⁻¹(18/24) = tan⁻¹(3/4).
8. (A) Let θ = sin⁻¹x. Then sin θ = x = x/1 (Opp/Hyp). Adjacent = √(1² - x²) = √(1-x²). cot θ = Adj/Opp = √(1-x²)/x.
9. (A) Use tan⁻¹x + tan⁻¹y = π + tan⁻¹((x+y)/(1-xy)) because xy = 23 = 6 > 1, and x>0, y>0. θ = π + tan⁻¹((2+3)/(1-23)) = π + tan⁻¹(5/-5) = π + tan⁻¹(-1) = π + (-π/4) = 3π/4. Then tan θ = tan(3π/4) = -1.
10. (C) sec⁻¹(-x) = π - sec⁻¹x. sec⁻¹(-2/√3) = π - sec⁻¹(2/√3). We know cos(π/6) = √3/2, so sec(π/6) = 2/√3. Thus sec⁻¹(2/√3) = π/6. The value is π - π/6 = 5π/6.

Study these notes carefully and practice more problems, especially from the NCERT Exemplar book. Good luck!