Trigonometric Identities

Introduction

Trigonometric identities are equations involving trigonometric ratios that hold true for all values of the angle. In ICSE Class 10, you learn the three fundamental identities and use them to prove more complex identities.

Fundamental Identities

  1. sin²A + cos²A = 1
  2. sec²A = 1 + tan²A
  3. cosec²A = 1 + cot²A

Derived Forms

From identity (1):

  • sin²A = 1 − cos²A
  • cos²A = 1 − sin²A

From identity (2):

  • tan²A = sec²A − 1
  • sec²A − tan²A = 1

From identity (3):

  • cot²A = cosec²A − 1
  • cosec²A − cot²A = 1

ICSE-Specific Angle Values

Memorise these values for ICSE problems:

Angle30°45°60°90°
sin0¹/₂¹/√2√3/21
cos1√3/2¹/√2¹/₂0
tan0¹/√31√3

Proving Trigonometric Identities — Strategy

Step 1: Choose the more complex side

Start with the side that looks more complicated (usually LHS).

Step 2: Convert to sine and cosine

Express all ratios in terms of sin and cos if simplification is needed.

Step 3: Use algebraic manipulation

Factorise, rationalise, or use the fundamental identities.

Step 4: Simplify step-by-step

Work towards the simpler side. Never cross-multiply across the equality sign.

Step 5: Verify the final expression matches

The simplified form must exactly match the other side.


Worked Examples

Example 1: Basic Identity Proof

Prove that: (1 − cos²A) cosec²A = 1

Solution: LHS = (1 − cos²A) × cosec²A = sin²A × cosec²A (since sin²A + cos²A = 1) = sin²A × 1/sin²A = 1 = RHS ✓

Example 2: Using sec²A − tan²A = 1

Prove that: sec⁴A − sec²A = tan⁴A + tan²A

Solution: LHS = sec⁴A − sec²A = sec²A(sec²A − 1) = sec²A × tan²A (since sec²A − 1 = tan²A) = tan²A × sec²A

RHS = tan⁴A + tan²A = tan²A(tan²A + 1) = tan²A × sec²A (since 1 + tan²A = sec²A)

LHS = RHS = tan²A × sec²A ✓

Example 3: Conversion to Sine and Cosine

Prove that: cotA + tanA = secA × cosecA

Solution: LHS = cosA/sinA + sinA/cosA = (cos²A + sin²A) / (sinA × cosA) = 1 / (sinA × cosA) = cosecA × secA = RHS ✓

Example 4: Using sin²A + cos²A = 1

Prove that: √[(1 − sinA) / (1 + sinA)] = secA − tanA

Solution: LHS = √[(1 − sinA) / (1 + sinA)] × √[(1 − sinA) / (1 − sinA)] = √[(1 − sinA)² / (1 − sin²A)] = √[(1 − sinA)² / cos²A] = (1 − sinA) / cosA = 1/cosA − sinA/cosA = secA − tanA = RHS ✓

Example 5: Using cosec²A − cot²A = 1

Prove that: (cosecA + cotA)(cosecA − cotA) = 1

Solution: LHS = (cosecA + cotA)(cosecA − cotA) = cosec²A − cot²A = 1 = RHS ✓


Comparison: The Three Identities

IdentityTrigonometric FormAlternative Form
sin²A + cos²A = 11 − sin²A = cos²A1 − cos²A = sin²A
sec²A − tan²A = 1tan²A = sec²A − 1sec²A = 1 + tan²A
cosec²A − cot²A = 1cot²A = cosec²A − 1cosec²A = 1 + cot²A

Common Mistakes and Fixes

MistakeFix
Writing (sinA)² as sinA²Always write sin²A for (sinA)²
Cancelling terms across the identityWork on one side only — never cross the = sign
Forgetting to rationalise the denominatorMultiply numerator and denominator by conjugate
Using identities in the wrong directionChoose the form that simplifies your expression
Incorrectly squaring trigonometric expressions(sinA + cosA)² = sin²A + cos²A + 2sinAcosA = 1 + 2sinAcosA

ICSE Exam Focus

Trigonometric identities carry 6–8 marks in ICSE exams. Questions require:

  • Proving identities using the three fundamental identities.
  • Expressing one ratio in terms of another.
  • Simplifying complicated trigonometric expressions.
  • Using algebraic techniques (factorisation, conjugates).

Marks Blueprint:

TopicMarks
Direct identity proof (simple)3
Complex identity proof5
Simplifying expressions2
Expressing one ratio in terms of another2

Self-Test Questions

  1. Prove that: √[(1 + cosA) / (1 − cosA)] = cosecA + cotA.

  2. Show that: (tanA + secA − 1) / (tanA − secA + 1) = (1 + sinA) / cosA.

  3. Prove that: sin⁴A − cos⁴A = sin²A − cos²A.

  4. If x = a sinθ + b cosθ and y = a cosθ − b sinθ, prove that x² + y² = a² + b².

  5. Prove that: (sinA − cosecA)² + (cosA − secA)² = cot²A + tan²A − 1.

  6. Prove that: (1 + tan²A) / (1 + cot²A) = tan²A.


In ICSE identity proofs, work on the LHS unless the RHS is clearly more complex. Avoid moving terms across the equality sign — manipulate one side until it matches the other.

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