StudyOSStudyOS CBSE Study Material, Class 5–12

Inverse Trigonometric Functions

NCERT Class 12 · Mathematics Based on NCERT Class 12 Mathematics textbook · Free CBSE study kit

Everything you need to ace this chapter, in one place.
Notes · Mind Map · Quiz · Assertion-Reason · Case Studies · Question Bank · Past Papers

Get Full Study Kit →
Studying with friends? Send them this chapter.

Chapter Notes

Overview
This chapter explores the inverse of trigonometric functions, which do not have inverses over their natural domains because they are not one-one and onto. By restricting the domains of sine, cosine, tangent, cosecant, secant, and cotangent to specific intervals, we define their principal value branches, making them invertible. The big idea is that inverse trigonometric functions, denoted by sin⁻¹, cos⁻¹, etc., are used to find angles from given trigonometric ratios, and they play a crucial role in calculus, science, and engineering. A student must understand the principal value branches, domains, and ranges to solve problems accurately.
Key Concepts (13)
Inverse of a function
If f: X→Y is one-one and onto, then a unique function g: Y→X exists such that g(y)=x, where x∈X and y=f(x). g is called the inverse of f and is denoted by f⁻¹.
e.g. If f(x)=2x for real numbers, then f⁻¹(y)=y/2, because f is one-one and onto.
Principal value branch
The branch of an inverse trigonometric function with the range that is the principal value branch, i.e., the specific interval where the function is one-one and onto and the inverse is defined.
e.g. For sin⁻¹, the principal value branch has range [-π/2, π/2]; for cos⁻¹, it is [0, π].
sin⁻¹ (arc sine function)
The inverse of the sine function restricted to the domain [-π/2, π/2], with domain [-1, 1] and range [-π/2, π/2].
e.g. sin⁻¹(1/2) = π/6 because sin(π/6)=1/2 and π/6 is in the principal range.
cos⁻¹ (arc cosine function)
The inverse of the cosine function restricted to the domain [0, π], with domain [-1, 1] and range [0, π].
e.g. cos⁻¹(√3/2) = π/6 because cos(π/6)=√3/2 and π/6 is in [0, π].
tan⁻¹ (arc tangent function)
The inverse of the tangent function restricted to the domain (-π/2, π/2), with domain R and range (-π/2, π/2).
e.g. tan⁻¹(1) = π/4 because tan(π/4)=1 and π/4 is in (-π/2, π/2).
cot⁻¹ (arc cotangent function)
The inverse of the cotangent function restricted to the domain (0, π), with domain R and range (0, π).
e.g. cot⁻¹(1/√3) = π/3 because cot(π/3)=1/√3 and π/3 is in (0, π).
sec⁻¹ (arc secant function)
The inverse of the secant function restricted to the domain [0, π] – {π/2}, with domain R – (-1, 1) and range [0, π] – {π/2}.
e.g. sec⁻¹(2) = π/3 because sec(π/3)=2 and π/3 is in the principal range.
cosec⁻¹ (arc cosecant function)
The inverse of the cosecant function restricted to the domain [-π/2, π/2] – {0}, with domain R – (-1, 1) and range [-π/2, π/2] – {0}.
e.g. cosec⁻¹(√2) = π/4 because cosec(π/4)=√2 and π/4 is in the principal range.
sin⁻¹x vs (sin x)⁻¹
sin⁻¹x is the inverse sine function (arc sine), not to be confused with (sin x)⁻¹ = 1/sin x, which is the reciprocal of sine.
e.g. sin⁻¹(0.5) = π/6, whereas (sin 30°)⁻¹ = 1/0.5 = 2.
Principal value
The value of an inverse trigonometric function that lies in the range of the principal branch is called the principal value.
e.g. The principal value of sin⁻¹(1/2) is π/6, not 5π/6, because π/6 is in [-π/2, π/2].
Graph of inverse function as mirror image
The graph of an inverse function can be obtained from the graph of the original function as a mirror image (reflection) along the line y = x.
e.g. The graph of y = sin⁻¹x is the reflection of the graph of y = sin x (restricted to [-π/2, π/2]) about the line y = x.
sin(sin⁻¹x) = x
For x ∈ [-1, 1], sin(sin⁻¹x) = x.
e.g. sin(sin⁻¹(0.5)) = sin(π/6) = 0.5.
sin⁻¹(sin x) = x
For x ∈ [-π/2, π/2], sin⁻¹(sin x) = x.
e.g. sin⁻¹(sin(π/4)) = π/4.
Formulas & Equations (18)
sin(sin⁻¹x) = x, x ∈ [-1, 1]
Reads as: sine of sine inverse x equals x, for x between negative one and one inclusive
Variables: x is a real number in the closed interval [-1, 1]
Use when: Simplifying expressions involving sin(sin⁻¹x) in problems or proving identities
⚠ Students forget the domain restriction and apply it for x outside [-1, 1].
sin⁻¹(sin x) = x, x ∈ [-π/2, π/2]
Reads as: sine inverse of sine x equals x, for x between negative pi by two and pi by two inclusive
Variables: x is an angle in radians in the closed interval [-π/2, π/2]
Use when: Finding the principal value of sin⁻¹(sin x) when x is in the principal range
⚠ Students apply this formula for x outside the principal range without adjusting the angle.
cos(cos⁻¹x) = x, x ∈ [-1, 1]
Reads as: cosine of cos inverse x equals x, for x between negative one and one inclusive
Variables: x is a real number in [-1, 1]
Use when: Simplifying expressions with cos(cos⁻¹x)
⚠ Forgetting the domain restriction.
cos⁻¹(cos x) = x, x ∈ [0, π]
Reads as: cos inverse of cos x equals x, for x between zero and pi inclusive
Variables: x is an angle in radians in [0, π]
Use when: Finding principal value of cos⁻¹(cos x) when x is in [0, π]
⚠ Applying for x outside [0, π] without conversion.
tan(tan⁻¹x) = x, x ∈ R
Reads as: tangent of tan inverse x equals x, for all real x
Variables: x is any real number
Use when: Simplifying tan(tan⁻¹x) expressions
⚠ None significant, but students may forget it applies for all real x.
tan⁻¹(tan x) = x, x ∈ (-π/2, π/2)
Reads as: tan inverse of tan x equals x, for x between negative pi by two and pi by two
Variables: x is an angle in radians in the open interval (-π/2, π/2)
Use when: Finding principal value of tan⁻¹(tan x) when x is in the principal range
⚠ Applying for x outside the interval without adjusting.
cot(cot⁻¹x) = x, x ∈ R
Reads as: cotangent of cot inverse x equals x, for all real x
Variables: x is any real number
Use when: Simplifying cot(cot⁻¹x)
⚠ None significant.
cot⁻¹(cot x) = x, x ∈ (0, π)
Reads as: cot inverse of cot x equals x, for x between zero and pi
Variables: x is an angle in radians in (0, π)
Use when: Finding principal value of cot⁻¹(cot x) when x is in (0, π)
⚠ Applying for x outside (0, π) without conversion.
sec(sec⁻¹x) = x, x ∈ R – (-1, 1)
Reads as: secant of sec inverse x equals x, for x in the set of real numbers excluding the open interval negative one to one
Variables: x is a real number with |x| ≥ 1
Use when: Simplifying sec(sec⁻¹x)
⚠ Applying for x in (-1, 1) where sec⁻¹x is not defined.
sec⁻¹(sec x) = x, x ∈ [0, π] – {π/2}
Reads as: sec inverse of sec x equals x, for x in the closed interval zero to pi excluding pi by two
Variables: x is an angle in radians in [0, π] except π/2
Use when: Finding principal value of sec⁻¹(sec x) when x is in the principal range
⚠ Forgetting to exclude π/2.
cosec(cosec⁻¹x) = x, x ∈ R – (-1, 1)
Reads as: cosecant of cosec inverse x equals x, for x in the set of real numbers excluding the open interval negative one to one
Variables: x is a real number with |x| ≥ 1
Use when: Simplifying cosec(cosec⁻¹x)
⚠ Applying for x in (-1, 1) where cosec⁻¹x is not defined.
cosec⁻¹(cosec x) = x, x ∈ [-π/2, π/2] – {0}
Reads as: cosec inverse of cosec x equals x, for x in the closed interval negative pi by two to pi by two excluding zero
Variables: x is an angle in radians in [-π/2, π/2] except 0
Use when: Finding principal value of cosec⁻¹(cosec x) when x is in the principal range
⚠ Forgetting to exclude 0.
sin⁻¹(2x√(1-x²)) = 2 sin⁻¹x, for -1/√2 ≤ x ≤ 1/√2
Reads as: sine inverse of 2x times square root of 1 minus x squared equals 2 sine inverse x, for x between negative one by root two and one by root two inclusive
Variables: x is a real number in [-1/√2, 1/√2]
Use when: Simplifying expressions or proving identities involving sin⁻¹(2x√(1-x²))
⚠ Applying the formula for x outside the specified domain without checking.
sin⁻¹(2x√(1-x²)) = 2 cos⁻¹x, for 1/√2 ≤ x ≤ 1
Reads as: sine inverse of 2x times square root of 1 minus x squared equals 2 cos inverse x, for x between one by root two and one inclusive
Variables: x is a real number in [1/√2, 1]
Use when: Simplifying expressions or proving identities with sin⁻¹(2x√(1-x²)) using cos substitution
⚠ Confusing the domain with the previous formula.
3 sin⁻¹x = sin⁻¹(3x - 4x³), for x ∈ [-1/2, 1/2]
Reads as: 3 sine inverse x equals sine inverse of 3x minus 4x cube, for x between negative one half and one half inclusive
Variables: x is a real number in [-1/2, 1/2]
Use when: Proving identities or simplifying triple angle inverse sine expressions
⚠ Forgetting the domain restriction and applying for all x.
3 cos⁻¹x = cos⁻¹(4x³ - 3x), for x ∈ [1/2, 1]
Reads as: 3 cos inverse x equals cos inverse of 4x cube minus 3x, for x between one half and one inclusive
Variables: x is a real number in [1/2, 1]
Use when: Proving identities or simplifying triple angle inverse cosine expressions
⚠ Applying the formula for x outside [1/2, 1].
tan⁻¹(√(1+x²)-1)/x, for x ≠ 0
Reads as: tan inverse of square root of 1 plus x squared minus 1 divided by x, for x not equal to zero
Variables: x is a non-zero real number
Use when: Simplifying expressions involving tan⁻¹ with square roots
⚠ Forgetting the condition x ≠ 0.
tan⁻¹(√(1-x²)/x), for |x| < a
Reads as: tan inverse of square root of a squared minus x squared divided by x, for absolute value of x less than a
Variables: x is a real number, a is a positive constant
Use when: Simplifying expressions with tan⁻¹ and square roots in trigonometric substitutions
⚠ Not considering the sign of x.
How It Works
Finding principal value of an inverse trigonometric function
  1. Step 1: Let the inverse function equal y, e.g., sin⁻¹x = y.
  2. Step 2: Rewrite as the trigonometric equation, e.g., sin y = x.
  3. Step 3: Identify the angle y in the principal value range of the function that satisfies the equation.
  4. Step 4: Ensure the angle is within the principal branch range (e.g., for sin⁻¹, y ∈ [-π/2, π/2]).
  5. Step 5: Write the principal value as the answer.
💡 Always check that the angle lies in the principal value branch range; otherwise, adjust using trigonometric identities.
Simplifying sin⁻¹(sin x) when x is not in the principal range
  1. Step 1: Identify the principal range of sin⁻¹, which is [-π/2, π/2].
  2. Step 2: If x is not in [-π/2, π/2], use trigonometric identities to rewrite sin x as sin of an angle that lies in the principal range.
  3. Step 3: For example, sin(3π/5) = sin(π - 3π/5) = sin(2π/5), and 2π/5 is in [-π/2, π/2].
  4. Step 4: Then sin⁻¹(sin x) = the angle from step 3 that is in the principal range.
💡 The formula sin⁻¹(sin x) = x holds only when x is in [-π/2, π/2]; otherwise, you must convert x to an equivalent angle in that range.
Proving identities like sin⁻¹(2x√(1-x²)) = 2 sin⁻¹x
  1. Step 1: Substitute x = sin θ or x = cos θ, depending on the domain.
  2. Step 2: Simplify the expression inside the inverse function using trigonometric identities (e.g., 2 sin θ cos θ = sin 2θ).
  3. Step 3: Apply the inverse function to the simplified trigonometric expression.
  4. Step 4: Ensure the resulting angle lies in the principal range of the inverse function.
  5. Step 5: Conclude the identity.
💡 The choice of substitution (sin or cos) depends on the domain of x given in the problem.
Writing tan⁻¹(cos x/(1 - sin x)) in simplest form
  1. Step 1: Use half-angle identities: cos x = cos²(x/2) - sin²(x/2) and 1 - sin x = (cos(x/2) - sin(x/2))².
  2. Step 2: Simplify the fraction to (cos(x/2) + sin(x/2))/(cos(x/2) - sin(x/2)).
  3. Step 3: Divide numerator and denominator by cos(x/2) to get (1 + tan(x/2))/(1 - tan(x/2)).
  4. Step 4: Recognize this as tan(π/4 + x/2).
  5. Step 5: Apply tan⁻¹ to get π/4 + x/2, ensuring the result is within the principal range of tan⁻¹.
💡 Always check the domain of x given in the problem to ensure the simplification is valid.
Exam Traps — Don't Make These Mistakes
⚠ Confusing sin⁻¹x with (sin x)⁻¹
✗ Students write sin⁻¹x = 1/sin x.
→ sin⁻¹x is the inverse sine function (arc sine), not the reciprocal. (sin x)⁻¹ = 1/sin x is the cosecant function.
✓ sin⁻¹x is the angle whose sine is x, and it is not equal to 1/sin x.
⚠ Applying sin⁻¹(sin x) = x for x outside the principal range
✗ sin⁻¹(sin(3π/5)) = 3π/5.
→ 3π/5 is not in the principal range of sin⁻¹, which is [-π/2, π/2]. The formula sin⁻¹(sin x) = x holds only when x is in that range.
✓ sin⁻¹(sin(3π/5)) = sin⁻¹(sin(π - 3π/5)) = sin⁻¹(sin(2π/5)) = 2π/5, since 2π/5 is in [-π/2, π/2].
⚠ Forgetting domain restrictions in identities like sin⁻¹(2x√(1-x²)) = 2 sin⁻¹x
✗ sin⁻¹(2x√(1-x²)) = 2 sin⁻¹x for all x in [-1, 1].
→ The identity holds only for x in [-1/√2, 1/√2]. For other x values, the right-hand side may not be in the principal range of sin⁻¹.
✓ sin⁻¹(2x√(1-x²)) = 2 sin⁻¹x, for -1/√2 ≤ x ≤ 1/√2.
⚠ Incorrect principal value for negative arguments
✗ The principal value of sin⁻¹(-1/2) is -π/6, but students sometimes write 7π/6 or 11π/6.
→ The principal value branch of sin⁻¹ is [-π/2, π/2], so the angle must be in this interval. 7π/6 and 11π/6 are not in this range.
✓ sin⁻¹(-1/2) = -π/6, because sin(-π/6) = -1/2 and -π/6 ∈ [-π/2, π/2].
⚠ Mixing up ranges of tan⁻¹ and cot⁻¹
✗ tan⁻¹x and cot⁻¹x have the same principal range.
→ tan⁻¹ has range (-π/2, π/2), while cot⁻¹ has range (0, π). They are different intervals.
✓ tan⁻¹: R → (-π/2, π/2); cot⁻¹: R → (0, π).
⚠ Forgetting to exclude points where the function is not defined
✗ sec⁻¹(sec(π/2)) = π/2.
→ sec x is not defined at x = π/2, so sec⁻¹(sec(π/2)) is not defined. The principal range of sec⁻¹ excludes π/2.
✓ sec⁻¹(sec x) is defined only for x ∈ [0, π] – {π/2}.
Compare & Contrast
Principal value range
sin⁻¹x
cos⁻¹x
sin⁻¹x has range [-π/2, π/2], while cos⁻¹x has range [0, π].
Principal value range
tan⁻¹x
cot⁻¹x
tan⁻¹x has range (-π/2, π/2), while cot⁻¹x has range (0, π).
Domain and range
sin⁻¹x
cosec⁻¹x
sin⁻¹x has domain [-1, 1] and range [-π/2, π/2]; cosec⁻¹x has domain R – (-1, 1) and range [-π/2, π/2] – {0}.
Domain and range
cos⁻¹x
sec⁻¹x
cos⁻¹x has domain [-1, 1] and range [0, π]; sec⁻¹x has domain R – (-1, 1) and range [0, π] – {π/2}.
Exam Focus
Marks pattern: 1-2 MCQs (1 mark each), 1 Very Short Answer (1 mark), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total around 8-10 marks.
Most asked: Finding principal values of inverse trigonometric functions · Simplifying expressions involving inverse trigonometric functions · Proving identities like sin⁻¹(2x√(1-x²)) = 2 sin⁻¹x · Evaluating sin⁻¹(sin x) or cos⁻¹(cos x) when x is not in the principal range · Solving equations involving inverse trigonometric functions
Don't skip: Proving identities using substitution (e.g., sin⁻¹(2x√(1-x²)) = 2 sin⁻¹x) — 3-5 marks; Simplifying and evaluating expressions like tan⁻¹(cos x/(1 - sin x)) — 3-5 marks; Solving equations like sin⁻¹(1-x) - 2 sin⁻¹x = π/2 — 5 marks
One-Line Revision (read aloud before the exam)
  1. sin⁻¹x is the angle whose sine is x, with principal range [-π/2, π/2].
  2. cos⁻¹x is the angle whose cosine is x, with principal range [0, π].
  3. tan⁻¹x is the angle whose tangent is x, with principal range (-π/2, π/2).
  4. cot⁻¹x is the angle whose cotangent is x, with principal range (0, π).
  5. sec⁻¹x is the angle whose secant is x, with principal range [0, π] – {π/2}.
  6. cosec⁻¹x is the angle whose cosecant is x, with principal range [-π/2, π/2] – {0}.
  7. sin⁻¹x is not equal to (sin x)⁻¹; (sin x)⁻¹ = 1/sin x.
  8. sin(sin⁻¹x) = x holds only for x ∈ [-1, 1].
  9. sin⁻¹(sin x) = x holds only for x ∈ [-π/2, π/2].
  10. For x outside the principal range, use identities like sin(π - x) = sin x to convert.
  11. The graph of an inverse function is the mirror image of the original function's graph about the line y = x.
  12. Principal value is the value of the inverse function that lies in its principal branch range.
  13. In identities like sin⁻¹(2x√(1-x²)) = 2 sin⁻¹x, always check the domain of x.
  14. tan⁻¹(1) = π/4, cos⁻¹(1/2) = π/3, sin⁻¹(√3/2) = π/3.
  15. cot⁻¹(-1/√3) = 2π/3 because cot(2π/3) = -1/√3 and 2π/3 ∈ (0, π).
  16. sec⁻¹(2/√3) = π/6 because sec(π/6) = 2/√3 and π/6 ∈ [0, π] – {π/2}.
  17. cosec⁻¹(-√2) = -π/4 because cosec(-π/4) = -√2 and -π/4 ∈ [-π/2, π/2] – {0}.
  18. The sum of sin⁻¹x and cos⁻¹x is π/2 for x ∈ [-1, 1].
  19. The sum of tan⁻¹x and cot⁻¹x is π/2 for x ∈ R.
  20. The sum of sec⁻¹x and cosec⁻¹x is π/2 for |x| ≥ 1.

MCQs — 10 Questions with Answers

Q1. What is the range of the function sin⁻¹ x?

  • A. [−π/2, π/2] ✓
  • B. [0, π]
  • C. (−π/2, π/2)
  • D. [−π, π]

Answer: A — By definition, the principal value branch of sin⁻¹: [−1, 1] → [−π/2, π/2] is a closed interval.

Q2. Which statement is correct regarding inverse trigonometric functions?

  • A. cos⁻¹(cos θ) = θ for all θ ∈ ℝ
  • B. sin⁻¹(sin θ) = θ only when θ ∈ [−π/2, π/2] ✓
  • C. tan⁻¹(tan θ) = θ for all θ ∈ ℝ
  • D. sec⁻¹ has domain [−1, 1]

Answer: B — The identity sin⁻¹(sin θ) = θ holds only when θ is in the principal value branch [−π/2, π/2]; outside this, we must first convert θ to an equivalent angle within the branch.

Q3. The domain of cosec⁻¹ is:

  • A. [−1, 1]
  • B. ℝ − (−1, 1) ✓
  • C. (−1, 1)
  • D. [−π/2, π/2]

Answer: B — Since cosec x = 1/sin x, the function takes all values except those in the open interval (−1, 1), so cosec⁻¹ has domain ℝ − (−1, 1).

Q4. Evaluate sin⁻¹(sin 7π/6).

  • A. 7π/6
  • B. −π/6 ✓
  • C. π/6
  • D. 5π/6

Answer: B — Since 7π/6 ∉ [−π/2, π/2], sin 7π/6 = −1/2; thus sin⁻¹(−1/2) = −π/6, which lies in the principal branch.

Q5. Which of the following is NOT true?

  • A. The range of tan⁻¹ is (−π/2, π/2)
  • B. The range of cos⁻¹ is [0, π]
  • C. The range of sec⁻¹ includes π/2 ✓
  • D. The domain of sin⁻¹ is [−1, 1]

Answer: C — The principal value branch of sec⁻¹ has range [0, π] − {π/2}, explicitly excluding π/2 because sec is undefined at this point.

Q6. If y = cos⁻¹ x, then which relation is correct?

  • A. cos y = x and y ∈ [−π/2, π/2]
  • B. cos y = x and y ∈ [0, π] ✓
  • C. sin y = x and y ∈ [0, π]
  • D. cos x = y and y ∈ [−1, 1]

Answer: B — By definition, if y = cos⁻¹ x, then cos y = x with y restricted to the principal value branch [0, π].

Q7. The graph of y = sin⁻¹ x is obtained from the graph of y = sin x by:

  • A. Reflection across the x-axis
  • B. Reflection across the line y = x ✓
  • C. Horizontal shift by π/2
  • D. Vertical stretch by factor 2

Answer: B — The inverse function's graph is the mirror image of the original function across the line y = x, obtained by swapping x and y coordinates.

Q8. For the function tan⁻¹: which statement is true?

  • A. Domain is [−1, 1] and range is [0, π]
  • B. Domain is ℝ and range is (−π/2, π/2) ✓
  • C. Domain is ℝ − {±1} and range is ℝ
  • D. Domain is [−π/2, π/2] and range is [−1, 1]

Answer: B — tan⁻¹: ℝ → (−π/2, π/2) has all real numbers as domain because the original tan function (when restricted) maps to all reals.

Q9. Which pair correctly matches the function to its principal value range? (Assertion-Reason style) Assertion: The range of cot⁻¹ is (0, π). Reason: cot is one-one and onto on the interval (0, π).

  • A. Both assertion and reason are true; reason explains assertion ✓
  • B. Both are true but reason does not explain assertion
  • C. Assertion true, reason false
  • D. Assertion false, reason true

Answer: A — cot⁻¹: ℝ → (0, π) because cot is one-one and onto when restricted to (0, π), making this the principal value branch.

Q10. Given that sin⁻¹ a + cos⁻¹ a = π/2 for a ∈ [−1, 1], find the value of sin⁻¹(1/2) + cos⁻¹(1/2). (HOTS)

  • A. π/6
  • B. π/3
  • C. π/2 ✓
  • D. 2π/3

Answer: C — Using the property sin⁻¹ x + cos⁻¹ x = π/2 for all x ∈ [−1, 1], substituting x = 1/2 gives sin⁻¹(1/2) + cos⁻¹(1/2) = π/2.

Flashcards

What is the principal value branch of sin⁻¹?

The domain is [−1, 1] and range is [−π/2, π/2], where sin⁻¹: [−1, 1] → [−π/2, π/2].

Define cos⁻¹ function with its domain and range.

cos⁻¹: [−1, 1] → [0, π], meaning for x ∈ [−1, 1], cos⁻¹ x gives a unique angle in [0, π].

What condition must θ satisfy for sin⁻¹(sin θ) = θ to hold?

θ must lie in the principal branch interval [−π/2, π/2]; otherwise sin⁻¹(sin θ) ≠ θ.

Why is the original sine function not one-one and onto?

Because sin x repeats every 2π and takes same value for different angles (e.g., sin π/6 = sin 5π/6), making it not injective; domain restriction to [−π/2, π/2] fixes this.

What is the domain and range of cosec⁻¹?

Domain is ℝ − (−1, 1) (all reals except open interval (−1, 1)) and principal range is [−π/2, π/2] − {0}.

How is the graph of y = sin⁻¹ x obtained from y = sin x?

By reflecting the graph of y = sin x across the line y = x, which is equivalent to interchanging x and y coordinates of each point.

Evaluate sin⁻¹(sin 3π/4). Explain why direct cancellation fails.

Since 3π/4 ∉ [−π/2, π/2], we cannot directly cancel; sin 3π/4 = √2/2 = sin π/4, so sin⁻¹(sin 3π/4) = π/4.

State the principal value branch of tan⁻¹ function.

tan⁻¹: ℝ → (−π/2, π/2), where domain is all real numbers and range is the open interval (−π/2, π/2).

If y = sin⁻¹ x, what is the relationship between x and y?

y = sin⁻¹ x means sin y = x, where x ∈ [−1, 1] and y ∈ [−π/2, π/2], establishing inverse relationship.

Why is the range of sec⁻¹ the interval [0, π] − {π/2} rather than all of [0, π]?

Because sec x = 1/cos x is undefined when cos x = 0 (i.e., at x = π/2), so sec⁻¹ must exclude π/2 from its range.

Important Board Questions

State the principal value branch of sin⁻¹ and write its domain and range. Give one example. [2 marks]

Define sin⁻¹: [−1, 1] → [−π/2, π/2] clearly; example: sin⁻¹(1/2) = π/6.

Why is the sine function not one-one and onto on ℝ, and how does domain restriction solve this problem? Derive the condition for sin⁻¹(sin θ) = θ. [5 marks]

Explain repetition (sin x repeats every 2π, same output for multiple angles); show sin is one-one and onto on [−π/2, π/2]; prove sin⁻¹(sin θ) = θ ⟺ θ ∈ [−π/2, π/2] using definition.

Prove that the graph of y = sin⁻¹ x is a reflection of y = sin x (restricted to [−π/2, π/2]) across the line y = x. Use this to explain why if (a, b) is on the sine graph, then (b, a) is on the sin⁻¹ graph. [6 marks]

Use inverse function property: if y = sin x with x ∈ [−π/2, π/2], then x = sin⁻¹ y; swapping variables gives y = sin⁻¹ x; geometrically, reflection across y = x swaps coordinates; verify with specific point like (π/6, 1/2) ↔ (1/2, π/6).

Assertion-Reason Questions for Inverse Trigonometric Functions

Q1. Assertion (A):

The function sin⁻¹: [-1, 1] → [-π/2, π/2] is bijective.

Reason (R):

The sine function restricted to [-π/2, π/2] is one-one and onto with range [-1, 1], so its inverse exists and is bijective.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why sin⁻¹ is bijective—it's the inverse of a bijective function (restricted sine). The existence of an inverse guarantees bijectivity of the inverse function.

Q2. Assertion (A):

cos⁻¹(x) is a decreasing function on its domain [-1, 1].

Reason (R):

The cosine function is increasing on its restricted domain [0, π], so its inverse must be decreasing.

Answer: C — A is true but R is false — The assertion is true (cos⁻¹ is indeed decreasing), but the reason is false. Cosine is actually decreasing on [0, π], not increasing. A decreasing original function produces a decreasing inverse, but the stated reason reverses this fact.

Q3. Assertion (A):

For x ∈ [-1, 1], we have sin(sin⁻¹(x)) = x.

Reason (R):

This follows from the definition that if y = sin⁻¹(x), then sin(y) = x.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason directly explains the assertion using the definition of inverse function. This is a fundamental property of inverse functions.

Unlock more Assertion-Reason questions in Premium — unlock →

Case-Based Questions (CBQ) for Inverse Trigonometric Functions

Case Study:

A telecommunications company designs an antenna that tracks satellites. The elevation angle θ from the ground to a satellite is given by θ = tan⁻¹(h/d), where h is the satellite's height above Earth and d is the horizontal distance to the base of the antenna. An engineer records that when h = 100 meters and d = 100 meters, θ₁ = tan⁻¹(1) = π/4. Later, the satellite moves such that h = 100√3 meters and d = 100 meters, making θ₂ = tan⁻¹(√3). The company needs to understand how the tracking angle changes with satellite position.

Q1. What is the value of θ₂ = tan⁻¹(√3)?

  • A. π/3 ✓
  • B. π/6
  • C. 2π/3
  • D. π/4

Answer: A

Q2. By how much does the elevation angle increase from θ₁ to θ₂?

  • A. π/12
  • B. π/6 ✓
  • C. π/4
  • D. π/3

Answer: B

Q3. If the engineer wants the elevation angle to be exactly π/6, what should be the ratio h:d?

  • A. 1:√3 ✓
  • B. √3:1
  • C. 1:1
  • D. 2:1

Answer: A

Q4. As the satellite's height h increases while d remains constant, the elevation angle θ:

  • A. Decreases toward 0
  • B. Increases toward π/2 ✓
  • C. Remains constant
  • D. Oscillates between π/4 and π/3

Answer: B

Unlock more case studies in Premium — unlock →

Previous Year Questions (PYQ) for Inverse Trigonometric Functions

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

If 2 cos–1x = y, then

  • 0 ≤ y ≤ π
  • –π ≤ y ≤ π
  • 0 ≤ y ≤ 2π
  • –π ≤ y ≤ 0

CBSE 2026 · Set 65/1/1 · 2 marks

Simplify: tan⁻¹((cos 2x + sin 2x)/(cos 2x – sin 2x)), 0 < x < π/4.

CBSE 2026 · Set 65/1/2 · 1 mark

If tan⁻¹ x = 3y, then

  • -π/2 < y < π/2
  • -3π/2 < y < 3π/2
  • -π/6 < y < π/6
  • -π/6 ≤ y ≤ π/6

CBSE 2026 · Set 65/1/2 · 2 marks

Simplify: tan⁻¹((cos 2x + sin 2x)/(cos 2x - sin 2x)), 0 < x < π/4.

CBSE 2026 · Set 65/1/2 · 2 marks

Simplify: cot⁻¹((1 + cos 2x)/(1 - cos 2x)), x ∈ [0, π/2].

CBSE 2026 · Set 65/1/3 · 1 mark

If sin⁻¹x + π/2 = y, then

  • -π/2 ≤ y ≤ π/2
  • -3π/2 ≤ y ≤ -π/2
  • π/2 ≤ y ≤ 3π/2
  • 0 ≤ y ≤ π

CBSE 2026 · Set 65/1/3 · 2 marks

Simplify: tan⁻¹[(cos 2x + sin 2x)/(cos 2x - sin 2x)], 0 < x < π/4

CBSE 2026 · Set 65/1/3 · 2 marks

Simplify: sin⁻¹(√(1 + cos 2x)/2), 0 < x < π/2

CBSE 2026 · Set 65/2/1 · 5 marks

Find the domain of g(x) = cos⁻¹ (x² - 1). Hence, find the value of x for which g(x) = π/3. Also, write the range of cos⁻¹ x other than its principal branch.

CBSE 2026 · Set 65/2/2 · 5 marks

Find the domain of p(x) = sin⁻¹(1 – 2x²). Hence, find the value of x for which p(x) = π/6. Also, write the range of 2p(x) + π/2.

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

If y = tan⁻¹[(1 – cos x)/sin x], then dy/dx is equal to :

  • 1
  • 2
  • 1/2
  • –1/2

CBSE 2026 · Set 65(B) R Mathematics · 3 marks

Find the value of sin⁻¹(3/5) + tan⁻¹(5/6) + cos⁻¹(13/6).

See all 67 board PYQs for this chapter — open PYQ tab →

Class 12 Maths Formula Sheet

Every formula from all 13 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

View the formula sheet →

Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

See all past board questions →

Next chapterMatrices →
Found this useful? Share it with your class group.

You've seen the free preview. Open the full interactive study kit — mind map, instant-graded quiz, progress tracking — free, no sign-in required.

Open Full Study Kit — Free →

Premium unlocks the full Question Bank, Daily Practice Papers, Chapter Quiz and Full Mock Papers for every chapter.