Introduction to Trigonometry Class 10 Notes (2026-27) — CBSE
Class 10 Maths Chapter 8 revision notes: all six trigonometric ratios, the standard-angle values table, the three identities with proofs, and exam question patterns.
Introduction to Trigonometry — Class 10 Maths Notes
Chapter Snapshot
Trigonometry studies the relationships between the sides and angles of a triangle. In Class 10 you work only with right-angled triangles and acute angles: you define six ratios from the triangle's sides, learn their exact values for the standard angles 0°, 30°, 45°, 60°, 90°, and prove/apply three identities that power almost every "prove that" question in the board exam.
Board relevance: one values-based evaluation question and one identity-proving question appear nearly every year. Together with Chapter 9 (Applications of Trigonometry), this unit is worth about 12 marks.
Syllabus note (rationalised): trigonometric ratios of complementary angles — sin(90°−θ) = cos θ etc. — has been removed from the current syllabus. Don't spend revision time there.
Key Concepts & Definitions
In a right triangle, for an acute angle θ:
- Opposite side (perpendicular) — the side facing θ
- Adjacent side (base) — the side next to θ (not the hypotenuse)
- Hypotenuse — the side opposite the right angle (always the longest)
The six trigonometric ratios:
Ratio Definition Reciprocal of
sin θ Opposite / Hypotenuse cosec θ
cos θ Adjacent / Hypotenuse sec θ
tan θ Opposite / Adjacent cot θ
cosec θ Hypotenuse / Opposite sin θ
sec θ Hypotenuse / Adjacent cos θ
cot θ Adjacent / Opposite tan θ
Memory aid: "Some People Have, Curly Brown Hair, Turned Permanently Black" → Sin = Perpendicular/Hypotenuse, Cos = Base/Hypotenuse, Tan = Perpendicular/Base.
Quotient relations:
- tan θ = sin θ / cos θ
- cot θ = cos θ / sin θ
Key facts:
- The value of sin θ or cos θ never exceeds 1 (the hypotenuse is the longest side).
- sec θ and cosec θ are always ≥ 1 for the angles where they are defined.
- tan θ can take any non-negative value for 0° ≤ θ < 90°.
- The ratios depend only on the angle, not on the size of the triangle (similar triangles give equal ratios).
- "sin θ" is one symbol — sin has no meaning without an angle, and sin²θ means (sin θ)², not sin(θ²).
Formulas
Values table for standard angles (memorise cold)
θ 0° 30° 45° 60° 90°
sin θ 0 1/2 1/√2 √3/2 1
cos θ 1 √3/2 1/√2 1/2 0
tan θ 0 1/√3 1 √3 not defined
cosec θ not defined 2 √2 2/√3 1
sec θ 1 2/√3 √2 2 not defined
cot θ not defined √3 1 1/√3 0
Table trick: under sin, write 0 1 2 3 4 → divide by 4 → square root → 0, 1/2, 1/√2, √3/2, 1. The cos row is the sin row reversed. tan = sin ÷ cos. The bottom three rows are reciprocals of the top three.
Why "not defined": tan θ = sin θ/cos θ, and cos 90° = 0 → division by zero. Same logic gives sec 90°, cot 0°, cosec 0° as not defined.
The three identities
Identity Valid for Rearrangements you'll use
sin²θ + cos²θ = 1 0° ≤ θ ≤ 90° sin²θ = 1 − cos²θ; cos²θ = 1 − sin²θ
1 + tan²θ = sec²θ 0° ≤ θ < 90° sec²θ − tan²θ = 1; (sec θ − tan θ)(sec θ + tan θ) = 1
1 + cot²θ = cosec²θ 0° < θ ≤ 90° cosec²θ − cot²θ = 1
Proof sketch (asked as a 2–3 marker): in right triangle ABC right-angled at B, Pythagoras gives AB² + BC² = AC². Divide throughout by AC² → (AB/AC)² + (BC/AC)² = 1 → cos²θ + sin²θ = 1. Dividing by AB² instead gives 1 + tan²θ = sec²θ; dividing by BC² gives cot²θ + 1 = cosec²θ.
Solved Patterns
Pattern 1 — evaluate using the table:
Evaluate 2 tan²45° + cos²30° − sin²60°.
= 2(1)² + (√3/2)² − (√3/2)² = 2.
Pattern 2 — one ratio given, find others:
Given tan θ = 8/15, find sin θ and cos θ.
Opposite = 8k, adjacent = 15k → hypotenuse = √(64k² + 225k²) = 17k.
sin θ = 8/17, cos θ = 15/17.
Pattern 3 — prove an identity:
Prove (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ).
LHS = (1/sin θ − cos θ/sin θ)² = (1 − cos θ)²/sin²θ = (1 − cos θ)²/(1 − cos²θ)
= (1 − cos θ)²/[(1 − cos θ)(1 + cos θ)] = (1 − cos θ)/(1 + cos θ) = RHS. ∎
Pattern 4 — find the angle:
If sin(A − B) = 1/2 and cos(A + B) = 1/2 with A B, find A and B.
A − B = 30° and A + B = 60° → A = 45°, B = 15°.
Important Question Patterns
1. Direct evaluation (1–2 marks): compute an expression from the values table. Watch for tan 90°/sec 90° traps — "not defined" makes the whole expression not defined.
2. One ratio → all ratios (2–3 marks): always draw the triangle, use Pythagoras with a scale factor k, then read off every ratio.
3. Prove the identity (3–4 marks): standard strategy — convert everything to sin and cos, take LCM, use sin²θ + cos²θ = 1, factorise using a² − b² where possible.
4. True/False & reasoning (1 mark): e.g. "sin θ = 4/3 is possible" — false, sin θ ≤ 1; "cot A is the product of cot and A" — false, cot without an angle is meaningless.
5. Angle-finding (2 marks): equations like sin(A−B), cos(A+B) given as table values; solve the pair of linear equations in A and B.
6. MCQ/assertion-reason (1 mark): identity rearrangements, e.g. the value of (sec A + tan A)(1 − sin A) simplifies to cos A.
⚡ Quick Revision
- SOH-CAH-TOA: sin = Opp/Hyp · cos = Adj/Hyp · tan = Opp/Adj; cosec, sec, cot are their reciprocals.
- Table trick: sin row = √(0,1,2,3,4)/2 → 0, 1/2, 1/√2, √3/2, 1; cos row = reverse; tan = sin/cos.
- Not defined: tan 90°, sec 90°, cot 0°, cosec 0°.
- Identities: sin²θ + cos²θ = 1 · 1 + tan²θ = sec²θ · 1 + cot²θ = cosec²θ (all from Pythagoras).
- sin θ and cos θ are never 1; sec θ, cosec θ never < 1; tan θ unbounded.
- Identity-proving strategy: everything → sin/cos → LCM → apply sin² + cos² = 1 → factorise.
- One-ratio problems: draw the triangle, sides as multiples of k, Pythagoras for the third side.
- Removed from syllabus: complementary angles sin(90°−θ) = cos θ — skip it.
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