Inverse Trigonometry

🟢 Lite — Quick Review

Rapid summary for last-minute revision before your exam.

Inverse Trigonometry — Key Facts for JEE Main sin⁻¹x (arcsin): range [−π/2, π/2]; cos⁻¹x (arccos): range [0, π]; tan⁻¹x (arctan): range (−π/2, π/2) Domain restrictions: sin⁻¹x, cos⁻¹x require x ∈ [−1, 1] Reciprocal functions: cosec⁻¹x (|x| ≥ 1), sec⁻¹x (|x| ≥ 1), cot⁻¹x (all real) Principal value branches are defined to make functions one-to-one ⚡ Exam tip: JEE Main often tests conversion between inverse trig forms — sin⁻¹x + cos⁻¹x = π/2 for x ∈ [−1, 1]!

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🟡 Standard — Core Study

Standard content for students with a few days to months.

Inverse Trigonometry — JEE Main Study Guide

Basic identities:

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

Important relations:

• sin⁻¹(−x) = −sin⁻¹x (odd function property)
• cos⁻¹(−x) = π − cos⁻¹x
• tan⁻¹(−x) = −tan⁻¹x (odd function)
• cot⁻¹(−x) = π − cot⁻¹x

Conversion formulas:

• tan⁻¹x = sin⁻¹(x/√(1+x²)) = cos⁻¹(1/√(1+x²))
• sin⁻¹x = tan⁻¹(x/√(1−x²)) for x ∈ (−1, 1)
• cos⁻¹x = tan⁻¹(√(1−x²)/x) for x ∈ (0, 1]

Sum formulas:

• tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)] when xy < 1
• tan⁻¹x + tan⁻¹y = π + tan⁻¹[(x+y)/(1−xy)] when xy > 1 and x, y > 0
• tan⁻¹x − tan⁻¹y = tan⁻¹[(x−y)/(1+xy)]
• 2 tan⁻¹x = tan⁻¹[2x/(1−x²)] for |x| < 1

Simplification patterns:

• sin⁻¹(sin θ) = θ only if θ ∈ [−π/2, π/2]; otherwise adjust
• cos⁻¹(cos θ) = θ only if θ ∈ [0, π]; otherwise adjust
• tan⁻¹(tan θ) = θ only if θ ∈ (−π/2, π/2); otherwise adjust

Graphs (sketch for reference):

• y = sin⁻¹x: passes through (0, 0), (−1, −π/2), (1, π/2)

• y = cos⁻¹x: passes through (0, π/2), (1, 0), (−1, π)

• y = tan⁻¹x: asymptotes at y = ±π/2, passes through (0, 0)

• Key formula: sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)] with appropriate branch

• Common trap: tan⁻¹x + tan⁻¹y formula has branch conditions — if 1 − xy = 0, the sum is π/2 or −π/2

• Exam weight: 1 question per year (4 marks); frequently combined with differentiation/integration

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🔴 Extended — Deep Dive

Comprehensive coverage for students on a longer study timeline.

Inverse Trigonometry — Comprehensive JEE Main Notes

tan⁻¹x + tan⁻¹y + tan⁻¹z = π when x + y + z = xyz This is a key identity for three-angle problems Proof: tan(A+B+C) formula; if tan(A+B+C) is undefined (denominator = 0), then A+B+C = π/2 mod π; with proper domain, it equals π

tan⁻¹x = cot⁻¹(1/x) for x > 0; for x < 0, there are sign adjustments

Double angle with inverse: 2 sin⁻¹x = sin⁻¹(2x√(1−x²)) for x ∈ [−1/√2, 1/√2]

Differentiation of inverse trig:

• d/dx (sin⁻¹x) = 1/√(1−x²)
• d/dx (cos⁻¹x) = −1/√(1−x²)
• d/dx (tan⁻¹x) = 1/(1+x²)
• d/dx (cot⁻¹x) = −1/(1+x²)
• d/dx (sec⁻¹x) = 1/(|x|√(x²−1))
• d/dx (cosec⁻¹x) = −1/(|x|√(x²−1))

Integration results:

• ∫ dx/√(1−x²) = sin⁻¹x + C
• ∫ dx/(1+x²) = tan⁻¹x + C
• ∫ dx/(x√(x²−1)) = sec⁻¹|x| + C

Solving equations with inverse trig: Example: sin⁻¹x + sin⁻¹(2x) = π/3 Method: sin⁻¹(2x) = π/3 − sin⁻¹x → take sin of both sides sin(π/3 − sin⁻¹x) = 2x sin(π/3)cos(sin⁻¹x) − cos(π/3)sin(sin⁻¹x) = 2x (√3/2)√(1−x²) − (1/2)x = 2x → solve for x

Understanding principal value ranges: The range of inverse trig functions is chosen so the function is one-to-one:

• sin: [−π/2, π/2] (monotonic increasing)
• cos: [0, π] (monotonic decreasing)
• tan: (−π/2, π/2) (monotonic increasing)

Simplification to standard angle: sin⁻¹(1/2) = π/6; sin⁻¹(√3/2) = π/3; sin⁻¹(−1/2) = −π/6 cos⁻¹(1/2) = π/3; cos⁻¹(√3/2) = π/6; cos⁻¹(−1/2) = 2π/3 tan⁻¹(1) = π/4; tan⁻¹(√3) = π/3; tan⁻¹(−√3) = −π/3

Three-angle identity proof: For A + B + C = π: tan A + tan B + tan C = tan A · tan B · tan C Proof: from tan(A+B+C) formula when denominator = 0

Real values of inverse expressions: sin⁻¹(sin 5π/3): since 5π/3 ∉ [−π/2, π/2], simplify: sin 5π/3 = −√3/2 → sin⁻¹(−√3/2) = −π/3 ✓

cot⁻¹x = π + tan⁻¹(1/x) for x > 0, = tan⁻¹(1/x) for x < 0 (branch convention varies) For JEE purposes, use the definition: cot⁻¹x ∈ (0, π)

Evaluating expressions: tan(cos⁻¹x) = √(1−x²)/x for x > 0 (positive square root, x in QI or QIV) sin(tan⁻¹x) = x/√(1+x²) cos(tan⁻¹x) = 1/√(1+x²)

• Remember: sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2; tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)]; for tan⁻¹x − tan⁻¹y = tan⁻¹[(x−y)/(1+xy)]; when x+y = xy, tan⁻¹x + tan⁻¹y + tan⁻¹z = π
• Previous years: “Find value of tan⁻¹(1/2) + tan⁻¹(1/3)” [2023]; “If sin⁻¹x + sin⁻¹(2x) = π/3, find x” [2024]; “Simplify: tan⁻¹[2sin(π/4)]” [2024]

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📊 JEE Main Exam Essentials

Detail	Value
Questions	90 (30 per subject)
Time	3 hours
Marks	300 (90 per subject)
Section	Physics (30), Chemistry (30), Mathematics (30)
Negative	−1 for wrong answer
Mode	Computer-based

🎯 High-Yield Topics for JEE Main Mathematics

• Calculus (Differentiation + Integration) — ~35 marks combined
• Coordinate Geometry (straight lines, circles, conics) — ~20 marks
• Algebra (Complex Numbers, Quadratics, P&C, Probability) — ~25 marks
• Trigonometry + Inverse Trigonometry — ~15 marks
• Vector + 3D — ~15 marks

📝 Previous Year Question Patterns

• Inverse Trigonometry: 1 question per year, 4 marks
• Common patterns: simplify sum of inverse trig, solve equations, evaluate tan(cos⁻¹x), prove identities
• Weight: medium frequency, high scoring if formulas are known

💡 Pro Tips

• Always use the principal value range — check if your result falls in the right range
• tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)] but add π if 1−xy < 0 and x, y > 0
• When simplifying sin(cos⁻¹x), draw a right triangle: cos θ = x → sin θ = √(1−x²)
• For tan(cot⁻¹x), use the reciprocal identity: cot⁻¹x = tan⁻¹(1/x) for x > 0
• Remember: tan⁻¹x ∈ (−π/2, π/2), so tan⁻¹x always lies between −90° and 90°
• 2 tan⁻¹x = tan⁻¹(2x/(1−x²)) works for |x| < 1; for |x| > 1, use the appropriate branch

🔗 Official Resources

• NTA Official JEE Main
• JEE Main Syllabus PDF

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