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NCERT Mathematics Class 12 - Chapter 2: Inverse Trigonometric Functions - Notes

CBSEClass 12Mathematicsप्रतिलोम त्रिकोणमितीय फलन

Learning Objectives

  • Understand inverse trigonometric functions and their domains/ranges
  • Learn principal value branches
  • Study important properties and identities
  • Solve problems using inverse trigonometric formulas

Key Concepts

Principal Value Branches

sin-1x: Domain [-1,1], Range [-π/2, π/2]. cos-1x: Domain [-1,1], Range [0, π]. tan-1x: Domain R, Range (-π/2, π/2). cot-1x: Domain R, Range (0, π). sec-1x: Domain R-(-1,1), Range [0,π]-{π/2}. cosec-1x: Domain R-(-1,1), Range [-π/2, π/2]-{0}.

Key Properties

Negative arguments: sin-1(-x) = -sin-1x. cos-1(-x) = π - cos-1x. tan-1(-x) = -tan-1x. cosec-1(-x) = -cosec-1x. sec-1(-x) = π - sec-1x. cot-1(-x) = π - cot-1x.

Reciprocal relations: sin-1(1/x) = cosec-1x. cos-1(1/x) = sec-1x. tan-1(1/x) = cot-1x (for x > 0).

Complementary relations: sin-1x + cos-1x = π/2. tan-1x + cot-1x = π/2. sec-1x + cosec-1x = π/2.

Important Formulas

tan-1x + tan-1y = tan-1[(x+y)/(1-xy)] when xy < 1.

tan-1x - tan-1y = tan-1[(x-y)/(1+xy)] when xy > -1.

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 |x| < 1.

3tan-1x = tan-1[(3x-x3)/(1-3x2)].

Conversion Between Inverse Trig Functions

sin-1x = cos-1√(1-x2) = tan-1[x/√(1-x2)] for 0 < x < 1. These conversions are essential for simplification problems in JEE.

Summary

Inverse trigonometric functions are defined by restricting domains of trig functions to make them bijective. Each has a specific principal value branch. Key properties include complementary relations, negative argument rules, and addition formulas for tan-1.

Important Terms

  • Principal value: The unique value in the defined range of the inverse function
  • Domain restriction: Limiting input to make inverse well-defined
  • Complementary pair: sin-1x + cos-1x = π/2

Quick Revision

  • sin-1x ∈ [-π/2, π/2]; cos-1x ∈ [0, π]; tan-1x ∈ (-π/2, π/2)
  • sin-1x + cos-1x = π/2; tan-1x + cot-1x = π/2
  • sin-1(-x) = -sin-1x (odd); cos-1(-x) = π - cos-1x
  • tan-1x + tan-1y = tan-1[(x+y)/(1-xy)] if xy < 1
  • 2tan-1x = sin-1(2x/(1+x2)) for |x| ≤ 1
  • tan-1(1) = π/4; sin-1(1) = π/2; cos-1(1) = 0
NCERT Mathematics Class 12 - Chapter 2: Inverse Trigonometric Functions - Notes | EduMunch