Introduction to Trigonometry Chapter 8 | Mathematics

Complete NCERT notes, trigonometric ratios (sin, cos, tan, cosec, sec, cot), values for specific angles (0°,30°,45°,60°,90°), trigonometric identities, and solved examples for CBSE Class 10 Board Exam 2025-26.

📖 Introduction to Trigonometry

Trigonometry (from Greek trigōnon "triangle" + metron "measure") is the study of relationships between angles and sides of a right-angled triangle.

📌 Right Triangle Naming

For angle θ in a right triangle:

  • Opposite side (Perpendicular): side opposite to angle θ
  • Adjacent side (Base): side adjacent to angle θ
  • Hypotenuse: longest side opposite to right angle
📐 Trigonometric Ratios

1️⃣ sin θ

sin θ = Opposite / Hypotenuse

2️⃣ cos θ

cos θ = Adjacent / Hypotenuse

3️⃣ tan θ

tan θ = Opposite / Adjacent = sinθ/cosθ

4️⃣ cosec θ

cosec θ = Hypotenuse / Opposite = 1/sinθ

5️⃣ sec θ

sec θ = Hypotenuse / Adjacent = 1/cosθ

6️⃣ cot θ

cot θ = Adjacent / Opposite = 1/tanθ = cosθ/sinθ
🔄 Reciprocal Relations
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
tan θ = sinθ/cosθ
cot θ = cosθ/sinθ
📊 Trigonometric Ratios for Specific Angles
Angle θ30°45°60°90°
sin θ01/21/√2√3/21
cos θ1√3/21/√21/20
tan θ01/√31√3Not defined
cosec θNot defined2√22/√31
sec θ12/√3√22Not defined
cot θNot defined√311/√30
💡 Easy Trick: For sin, remember 0°=√0/2, 30°=√1/2, 45°=√2/2, 60°=√3/2, 90°=√4/2. Cos is reverse order.
🔑 Trigonometric Identities

📌 Fundamental Identities

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ

📝 Derived Forms

sin²θ = 1 - cos²θ
cos²θ = 1 - sin²θ
tan²θ = sec²θ - 1
cot²θ = cosec²θ - 1
Trigonometric Ratios of Complementary Angles
sin(90° - θ) = cos θ
cos(90° - θ) = sin θ
tan(90° - θ) = cot θ
cot(90° - θ) = tan θ
sec(90° - θ) = cosec θ
cosec(90° - θ) = sec θ
📐 All Important Formulas

1️⃣ Basic Ratios

sinθ = opp/hyp, cosθ = adj/hyp, tanθ = opp/adj

2️⃣ Reciprocal Ratios

cosecθ=1/sinθ, secθ=1/cosθ, cotθ=1/tanθ

3️⃣ Quotient Relations

tanθ = sinθ/cosθ, cotθ = cosθ/sinθ

4️⃣ Pythagorean Identities

sin²θ+cos²θ=1, 1+tan²θ=sec²θ, 1+cot²θ=cosec²θ

5️⃣ Complementary Angles

sin(90°-θ)=cosθ, tan(90°-θ)=cotθ, etc.

6️⃣ Special Values

sin45°=1/√2, cos60°=1/2, tan30°=1/√3, etc.
📝 NCERT Solved Examples

Example 1: Finding Ratios

If sin A = 3/4, find cos A and tan A.
cos A = √(1 - sin²A) = √(1 - 9/16) = √(7/16) = √7/4
tan A = sinA/cosA = (3/4)/(√7/4) = 3/√7

Example 2: Value of Expression

Evaluate: sin²30° + cos²60° - tan²45°
= (1/2)² + (1/2)² - (1)² = 1/4 + 1/4 - 1 = 1/2 - 1 = -1/2

Example 3: Using Identity

Prove: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A
LHS = sin²A + cosec²A + 2 + cos²A + sec²A + 2 = (sin²A+cos²A) + (cosec²A+sec²A) + 4
= 1 + (1+cot²A)+(1+tan²A)+4 = 7 + tan²A + cot²A = RHS

Example 4: Complementary Angles

Evaluate: sin 50°/cos 40° + tan 70°/cot 20°
sin50° = cos40°, tan70° = cot20°
= 1 + 1 = 2
Key Points for Board Exam

📌 Quick Reference Table

RatioAbbreviationFormula
SinesinOpposite/Hypotenuse
CosinecosAdjacent/Hypotenuse
TangenttanOpposite/Adjacent
Cosecantcosec1/sin
Secantsec1/cos
Cotangentcot1/tan
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