Inverse Trigonometric Functions — Mathematics Class 12 Notes (CBSE & HBSE)
Free NCERT Mathematics notes for Inverse Trigonometric Functions (Class 12) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 2 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.
Board exam focus — Inverse Trigonometric Functions (CBSE & HBSE)
Inverse Trigonometric Functions carries 5-8 marks in CBSE. Students must know the principal value branch domains/ranges for all six inverse trig functions and be able to simplify expressions using standard identities.
Principal Value Branches and Domain-Range
Inverse Trigonometric Functions — Principal Values
Inverse trigonometric functions are defined by restricting the domain of trigonometric functions so that they become one-one and onto.
| Function | Domain | Principal Value Branch (Range) |
|---|---|---|
| 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) | ℝ−(−1,1) | [0,π]−{π/2} |
| cosec⁻¹(x) | ℝ−(−1,1) | [−π/2,π/2]−{0} |
How to Find Principal Values
Rule: sin⁻¹(x) = θ means sinθ = x AND θ ∈ [−π/2, π/2].
Worked Example: Find sin⁻¹(−√3/2)
- We need θ ∈ [−π/2, π/2] such that sinθ = −√3/2
- sin(π/3) = √3/2, and since the value is negative, θ = −π/3
- ∴ sin⁻¹(−√3/2) = −π/3
Worked Example: Find cos⁻¹(−1/2)
- We need θ ∈ [0, π] such that cosθ = −1/2
- cos(π/3) = 1/2, and cosine is negative in second quadrant
- θ = π − π/3 = 2π/3
Key Values Table
| θ | 0 | π/6 | π/4 | π/3 | π/2 |
|---|---|---|---|---|---|
| sinθ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cosθ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tanθ | 0 | 1/√3 | 1 | √3 | ∞ |
CBSE Tip: Always write the final answer in the principal value branch. If you get an angle outside the range, convert it using the appropriate property.
Properties and Identities of Inverse Trig Functions
Standard Properties and Identities
Property 1: Negative Arguments
- sin⁻¹(−x) = −sin⁻¹(x) for x ∈ [−1,1]
- tan⁻¹(−x) = −tan⁻¹(x) for x ∈ ℝ
- cosec⁻¹(−x) = −cosec⁻¹(x) for |x| ≥ 1
- cos⁻¹(−x) = π − cos⁻¹(x) for x ∈ [−1,1]
- sec⁻¹(−x) = π − sec⁻¹(x) for |x| ≥ 1
- cot⁻¹(−x) = π − cot⁻¹(x) for x ∈ ℝ
Property 2: Complementary Angles
- sin⁻¹(x) + cos⁻¹(x) = π/2 for x ∈ [−1,1]
- tan⁻¹(x) + cot⁻¹(x) = π/2 for x ∈ ℝ
- sec⁻¹(x) + cosec⁻¹(x) = π/2 for |x| ≥ 1
Property 3: Reciprocal Relations
- sin⁻¹(1/x) = cosec⁻¹(x) for |x| ≥ 1
- cos⁻¹(1/x) = sec⁻¹(x) for |x| ≥ 1
- tan⁻¹(1/x) = cot⁻¹(x) for x > 0
- tan⁻¹(1/x) = −π + cot⁻¹(x) for x < 0
Property 4: tan⁻¹ Addition/Subtraction Formulas
- 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
- tan⁻¹x − tan⁻¹y = tan⁻¹[(x−y)/(1+xy)] if xy > −1
Property 5: Double Angle Formulas
- 2sin⁻¹x = sin⁻¹(2x√(1−x²))
- 2cos⁻¹x = cos⁻¹(2x²−1)
- 2tan⁻¹x = sin⁻¹[2x/(1+x²)] = tan⁻¹[2x/(1−x²)] = cos⁻¹[(1−x²)/(1+x²)]
Simplification Strategy
For expressions like sin(2cos⁻¹x): let θ = cos⁻¹x, then cosθ = x, and sin 2θ = 2sinθ cosθ = 2x√(1−x²).
CBSE Tip: The identity sin⁻¹x + cos⁻¹x = π/2 is extremely useful for simplification. Memorize all three complementary pairs.
Frequently asked questions
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Do these notes follow CBSE and HBSE?
Yes. The Inverse Trigonometric Functions notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.
What does the Inverse Trigonometric Functions chapter cover?
Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Inverse Trigonometric Functions.