Chapter 2 — Inverse Trigonometric Functions — covers domains, ranges, properties, and simplification of inverse trig expressions. Carries 6-8 marks. Master the principal value branches and key properties.
Key Concepts
Principal Value Branches
| Function | Domain | Range (Principal Value) |
|---|---|---|
| sin⁻¹x | [−1, 1] | [−π/2, π/2] |
| cos⁻¹x | [−1, 1] | [0, π] |
| tan⁻¹x | R (all reals) | (−π/2, π/2) |
| cot⁻¹x | R | (0, π) |
| sec⁻¹x | R − (−1, 1) | [0, π] − {π/2} |
| cosec⁻¹x | R − (−1, 1) | [−π/2, π/2] − {0} |
Important Properties
sin⁻¹x + cos⁻¹x = π/2
tan⁻¹x + cot⁻¹x = π/2
sec⁻¹x + cosec⁻¹x = π/2
Negative arguments:
sin⁻¹(−x) = −sin⁻¹x
cos⁻¹(−x) = π − cos⁻¹x
tan⁻¹(−x) = −tan⁻¹x
Reciprocal:
sin⁻¹(1/x) = cosec⁻¹x
cos⁻¹(1/x) = sec⁻¹x
tan⁻¹(1/x) = cot⁻¹x (for x > 0)
Sum formulas:
tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)), if xy < 1
tan⁻¹x − tan⁻¹y = tan⁻¹((x−y)/(1+xy))
Double angle:
2tan⁻¹x = sin⁻¹(2x/(1+x²)) = cos⁻¹((1−x²)/(1+x²)) = tan⁻¹(2x/(1−x²))
Solved Examples
Example 1
Q: Find the principal value of sin⁻¹(−1/2).
Solution: sin⁻¹(−1/2) = −sin⁻¹(1/2) = −π/6
(Range of sin⁻¹ is [−π/2, π/2], and sin(π/6) = 1/2)
Example 2
Q: Prove that tan⁻¹(1/2) + tan⁻¹(1/3) = π/4
Solution: Using tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy))
= tan⁻¹((1/2 + 1/3)/(1 − 1/6)) = tan⁻¹((5/6)/(5/6)) = tan⁻¹(1) = π/4 ✓
Example 3
Q: Simplify: cos⁻¹(1/2) + 2sin⁻¹(1/2)
Solution: cos⁻¹(1/2) = π/3; sin⁻¹(1/2) = π/6
= π/3 + 2(π/6) = π/3 + π/3 = 2π/3
Quick Revision Points
- sin⁻¹x ∈ [−π/2, π/2]; cos⁻¹x ∈ [0, π]; tan⁻¹x ∈ (−π/2, π/2)
- sin⁻¹x + cos⁻¹x = π/2 (always!)
- sin⁻¹(−x) = −sin⁻¹x; cos⁻¹(−x) = π − cos⁻¹x
- tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) when xy < 1
- 2tan⁻¹x = sin⁻¹(2x/(1+x²)) for |x| ≤ 1
- Always check if answer falls within principal value range!
Chapter Navigation
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Related Chapters in Class 12 Maths
- Relations and Functions Class 12 Notes
- Continuity and Differentiability Class 12 Notes
- Matrices Class 12 Notes
Practice What You Learned
Test yourself with our JEE Main Mock Test Set 1 to see how well you’ve mastered the concepts.