Trigonometric Ratios
In a right-angled triangle ABC with right angle at B:
For angle θ at vertex A:
- Opposite side = BC (side opposite to θ)
- Adjacent side = AB (side next to θ)
- Hypotenuse = AC (side opposite to 90°)
| Ratio | Formula | Reciprocal |
|---|---|---|
| sin θ | Opposite / Hypotenuse | cosec θ = 1/sin θ |
| cos θ | Adjacent / Hypotenuse | sec θ = 1/cos θ |
| tan θ | Opposite / Adjacent | cot θ = 1/tan θ |
Also: tan θ = sin θ / cos θ
Memory trick: "Some People Have Curly Brown Hair Through Proper Brushing"
- Sin = Perpendicular / Hypotenuse
- Cos = Base / Hypotenuse
- Tan = Perpendicular / Base
Standard Values Table
| Angle | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | undefined |
| cosec | undefined | 2 | √2 | 2/√3 | 1 |
| sec | 1 | 2/√3 | √2 | 2 | undefined |
| cot | undefined | √3 | 1 | 1/√3 | 0 |
Memory trick for sin row: Think of √0/2, √1/2, √2/2, √3/2, √4/2 = 0, 1/2, 1/√2, √3/2, 1
cos row = reverse of sin row; tan row = sin/cos
Finding Trigonometric Ratios from Given Information
Method: Draw a right triangle, use Pythagoras to find the missing side, then write all ratios.
Example: If sin A = 3/5, find all other trigonometric ratios of A.
sin A = 3/5 means opposite = 3, hypotenuse = 5. By Pythagoras: adjacent = √(25−9) = √16 = 4
So: cos A = 4/5, tan A = 3/4, cosec A = 5/3, sec A = 5/4, cot A = 4/3
Trigonometric Identities
The three fundamental identities:
$$\sin^2\theta + \cos^2\theta = 1$$ $$1 + \tan^2\theta = \sec^2\theta$$ $$1 + \cot^2\theta = \text{cosec}^2\theta$$
Derived forms (very useful for proofs):
- sin²θ = 1 − cos²θ; cos²θ = 1 − sin²θ
- tan²θ = sec²θ − 1; sec²θ − tan²θ = 1
- cot²θ = cosec²θ − 1; cosec²θ − cot²θ = 1
- (sec θ + tan θ)(sec θ − tan θ) = 1
- (cosec θ + cot θ)(cosec θ − cot θ) = 1
Worked proof example: Prove: (1 + tan²A)/(1 + cot²A) = tan²A
LHS = sec²A / cosec²A = (1/cos²A) / (1/sin²A) = sin²A/cos²A = tan²A = RHS ✓
Complementary Angles
Two angles are complementary if they add up to 90°. In a right triangle, the two acute angles are always complementary.
$$\sin(90° - \theta) = \cos\theta$$ $$\cos(90° - \theta) = \sin\theta$$ $$\tan(90° - \theta) = \cot\theta$$ $$\cot(90° - \theta) = \tan\theta$$ $$\sec(90° - \theta) = \text{cosec}\theta$$ $$\text{cosec}(90° - \theta) = \sec\theta$$
Worked examples:
Q1. Evaluate: sin 18° / cos 72°
cos 72° = cos(90°−18°) = sin 18° So sin 18° / sin 18° = 1
Q2. Evaluate: tan 48° × tan 16° × tan 42° × tan 74°
tan 42° = tan(90°−48°) = cot 48°; tan 74° = cot 16° = tan 48° × tan 16° × cot 48° × cot 16° = (tan 48° × cot 48°) × (tan 16° × cot 16°) = 1 × 1 = 1
Worked Trigonometric Proofs
Q. Prove: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A
LHS = sin²A + 2 + cosec²A + cos²A + 2 + sec²A = (sin²A + cos²A) + (1 + cosec²A) + (1 + sec²A) + 2 = 1 + (1 + cot²A) + (1 + tan²A) + 2 = 1 + 1 + cot²A + 1 + tan²A + 2 = 5 + tan²A + cot²A
Hmm that gives 5 not 7 — let's redo: = sin²A + 2sinAcosecA + cosec²A + cos²A + 2cosAsecA + sec²A = (sin²A + cos²A) + 2(sinA × 1/sinA) + 2(cosA × 1/cosA) + cosec²A + sec²A = 1 + 2 + 2 + (1 + cot²A) + (1 + tan²A) = 1 + 2 + 2 + 1 + cot²A + 1 + tan²A = 7 + tan²A + cot²A ✓
Board Exam Tips
- 8–10 marks from Chapters 8 and 9 combined — largest Physics chunk
- Memorise the table for 0°, 30°, 45°, 60°, 90° — many marks depend on it
- For proofs: start with the more complex side; express everything in sin and cos if stuck
- Complementary angle properties reduce expressions quickly — look for pairs adding to 90°
- Identity proofs strategy: Use sin²+cos²=1 as starting point, factor strategically
- tan = sin/cos substitution often simplifies complex expressions
- Don't confuse sin²θ with sin(θ²) — the exponent is on the trig ratio, not the angle
📚 RD Sharma Mathematics Class 10 — Trigonometry with Proofs
📝 Oswaal CBSE Maths Question Bank — Trigonometry Board Questions