1. Chapter at a Glance
The word trigonometry means "measurement of triangles". Indian astronomers made huge contributions here: Āryabhaṭa (5th century) gave tables of jyā (half-chords), and the word "sine" itself comes, through Arabic and Latin translations, from the Sanskrit jyā/jīvā. In this chapter we define six trigonometric ratios of an acute angle in a right triangle, find their values for 0°, 30°, 45°, 60° and 90°, and prove and use the identity sin²A + cos²A = 1. Trigonometry (this chapter plus the next) carries about 12 marks in the CBSE board paper.
- Sides of a right triangle relative to an angle: opposite, adjacent, hypotenuse.
- The six ratios: sin, cos, tan, cosec, sec, cot — and why they depend only on the angle.
- Finding all ratios when one is given.
- Values at 0°, 30°, 45°, 60°, 90°.
- Identities: sin²A + cos²A = 1, 1 + tan²A = sec²A, 1 + cot²A = cosec²A.
- Proving simple identities.
2. Trigonometric Ratios
| Ratio | Definition | Reciprocal of |
|---|---|---|
| sin A | opposite / hypotenuse = BC/AC | cosec A |
| cos A | adjacent / hypotenuse = AB/AC | sec A |
| tan A | opposite / adjacent = BC/AB | cot A |
| cosec A | hypotenuse / opposite | sin A |
| sec A | hypotenuse / adjacent | cos A |
| cot A | adjacent / opposite | tan A |
Also tan A = sin A / cos A and cot A = cos A / sin A.
Why ratios depend only on the angle: any two right triangles with the same acute angle A are similar (AA), so the ratios of their corresponding sides are equal.
For an acute angle, 0 < sin A < 1 and 0 < cos A < 1, because the hypotenuse is the longest side. "sin A" is a single symbol — it is not "sin × A".
If tan A = 4/3, find sin A and cos A.
Solution: Opposite = 4k, adjacent = 3k, hypotenuse = 5k. sin A = 4/5, cos A = 3/5.
If 15 cot A = 8, find sin A and sec A.
Solution: cot A = 8/15 → adjacent 8k, opposite 15k, hypotenuse 17k. sin A = 15/17, sec A = 17/8.
3. Ratios of Specific Angles
45°: an isosceles right triangle with legs 1, 1 has hypotenuse √2 → sin 45° = cos 45° = 1/√2, tan 45° = 1.
30° and 60°: halve an equilateral triangle of side 2: the halves have sides 1, √3, 2 with angles 30°, 60°, 90°.
| ∠A | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin A | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos A | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan A | 0 | 1/√3 | 1 | √3 | not defined |
| cosec A | not defined | 2 | √2 | 2/√3 | 1 |
| sec A | 1 | 2/√3 | √2 | 2 | not defined |
| cot A | not defined | √3 | 1 | 1/√3 | 0 |
sin of 0°, 30°, 45°, 60°, 90° = √0/2, √1/2, √2/2, √3/2, √4/2. Cos is the same list reversed. As A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0.
Chart — sin A as A increases:
Evaluate 2 tan²45° + cos²30° − sin²60°.
Solution: 2(1) + 3/4 − 3/4 = 2.
If sin(A − B) = 1/2 and cos(A + B) = 1/2 (0° < A + B ≤ 90°, A > B), find A and B.
Solution: A − B = 30°, A + B = 60° → A = 45°, B = 15°.
4. Trigonometric Identities
In right △ABC (right angle at B): AB² + BC² = AC². Divide by AC²:
- sin²A + cos²A = 1
- 1 + tan²A = sec²A (divide by AB²; A ≠ 90°)
- 1 + cot²A = cosec²A (divide by BC²; A ≠ 0°)
These let us express every ratio in terms of any one of them, e.g. cos A = √(1 − sin²A), tan A = sin A/√(1 − sin²A).
Prove (1 + tan²A)/(1 + cot²A) = tan²A.
Proof: LHS = sec²A / cosec²A = (1/cos²A) × sin²A = sin²A/cos²A = tan²A = RHS. ∎
Prove (sec A − tan A)(sec A + tan A) = 1, and hence find sec A − tan A if sec A + tan A = 3.
Solution: LHS = sec²A − tan²A = 1. So sec A − tan A = 1/3.
5. Solved Problems (Board Pattern)
In △PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find sin P, cos P and tan P.
Solution: Let QR = x, PR = 25 − x. 25 + x² = (25 − x)² → 25 + x² = 625 − 50x + x² → x = 12. PR = 13. sin P = 12/13, cos P = 5/13, tan P = 12/5.
Prove: (sin θ − cos θ + 1)/(sin θ + cos θ − 1) = 1/(sec θ − tan θ).
Proof: Divide numerator and denominator by cos θ: LHS = (tan θ + sec θ − 1)/(tan θ − sec θ + 1).
In the numerator, replace 1 by sec²θ − tan²θ = (sec θ − tan θ)(sec θ + tan θ): numerator = (tan θ + sec θ)[1 − (sec θ − tan θ)] = (tan θ + sec θ)(tan θ − sec θ + 1).
The second bracket equals the denominator, so LHS = tan θ + sec θ = 1/(sec θ − tan θ), since (sec θ − tan θ)(sec θ + tan θ) = 1. ∎
Express sin A, sec A and tan A in terms of cot A.
Solution: cosec²A = 1 + cot²A → sin A = 1/√(1 + cot²A). tan A = 1/cot A. sec A = √(1 + tan²A) = √(1 + cot²A)/cot A.
6. Master Formula Table
| Relation | Formula |
|---|---|
| Quotient | tan A = sin A/cos A; cot A = cos A/sin A |
| Reciprocal | cosec A = 1/sin A; sec A = 1/cos A; cot A = 1/tan A |
| Pythagorean I | sin²A + cos²A = 1 |
| Pythagorean II | sec²A − tan²A = 1 |
| Pythagorean III | cosec²A − cot²A = 1 |
| Key values | sin 30° = ½, cos 60° = ½, tan 45° = 1, sin 60° = √3/2 |
7. Master Summary Table
| Topic | Key idea | Exam focus |
|---|---|---|
| Ratios | Side ratios relative to angle | Find all ratios from one |
| Standard angles | 45-45-90 and 30-60-90 triangles | Evaluate expressions |
| Find angles | Match values | sin(A − B) = ½ type |
| Identities | From Pythagoras | Prove identities (3 marks) |
| Ranges | 0 ≤ sin, cos ≤ 1 | True/false MCQs |
8. Mind Map
- sin, cos, tan
- Reciprocals
- 0°, 30°, 45°
- 60°, 90°
- sin² + cos² = 1
- sec² − tan² = 1
- Āryabhaṭa's jyā
- "sine" from jyā
9. Tips and Tricks
- To prove identities, convert everything to sin and cos when stuck.
- Work on the more complicated side and simplify it to the other side.
- Remember "Some People Have Curly Brown Hair Through Proper Brushing": sin = P/H, cos = B/H, tan = P/B (P = perpendicular/opposite, B = base, H = hypotenuse).
- sin(A + B) ≠ sin A + sin B (e.g. sin 90° = 1, but sin 60° + sin 30° ≈ 1.37).
- Writing sin²A as sin A².
- Saying tan 90° = ∞ in an answer — write "not defined".
10. Important One-liners
- The value of sin A or cos A never exceeds 1 for an acute angle.
- tan 45° = 1 and sin 30° = cos 60° = ½.
- sec A and cosec A are always ≥ 1 for acute angles.
- sin²A + cos²A = 1 for every angle A.
- Trigonometric ratios depend only on the angle, not on the size of the triangle.
11. Exam-pattern Questions with Detailed Answers
The value of (sin 30° + cos 60°) is:
- 0
- 1
- √3
- 1/2
Show answer
½ + ½ = 1.
Answer: (B)
If sin A = 3/4, find cos A and tan A.
Show answer
Adjacent = √(16 − 9) = √7. cos A = √7/4, tan A = 3/√7.
Answer: √7/4 and 3/√7
Evaluate sin 60° cos 30° + sin 30° cos 60°.
Show answer
(√3/2)(√3/2) + (½)(½) = ¾ + ¼ = 1.
Answer: 1
Evaluate (5 cos²60° + 4 sec²30° − tan²45°)/(sin²30° + cos²30°).
Show answer
Numerator: 5/4 + 16/3 − 1 = (15 + 64 − 12)/12 = 67/12. Denominator = 1. Value = 67/12.
Answer: 67/12
If √3 tan θ = 1, find sin²θ − cos²θ.
Show answer
tan θ = 1/√3 → θ = 30°. ¼ − ¾ = −½.
Answer: −1/2
Prove: (1 − cos²A) cosec²A = 1.
Show answer
LHS = sin²A × 1/sin²A = 1. ∎
Answer: Proved
State true or false: "cos A is the abbreviation for cosecant of angle A." Explain.
Show answer
False. cos A is cosine; cosecant is written cosec A.
Answer: False
If tan(A + B) = √3 and tan(A − B) = 1/√3, 0° < A + B ≤ 90°, A > B, find A and B.
Show answer
A + B = 60°, A − B = 30° → A = 45°, B = 15°.
Answer: A = 45°, B = 15°
Prove: (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A.
Show answer
Divide numerator and denominator by sin A: (cot A − 1 + cosec A)/(cot A + 1 − cosec A). Write 1 = cosec²A − cot²A in the numerator: (cot A + cosec A) − (cosec²A − cot²A) = (cot A + cosec A)(1 − cosec A + cot A). The bracket equals the denominator, so LHS = cot A + cosec A. ∎
Answer: Proved
Prove: √[(1 + sin A)/(1 − sin A)] = sec A + tan A.
Show answer
Multiply inside by (1 + sin A)/(1 + sin A): √[(1 + sin A)²/(1 − sin²A)] = (1 + sin A)/cos A = sec A + tan A. ∎
Answer: Proved
If sin θ + cos θ = √3, prove that tan θ + cot θ = 1.
Show answer
Square: 1 + 2 sin θ cos θ = 3 → sin θ cos θ = 1. tan θ + cot θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/1 = 1. ∎
Answer: Proved
Prove: (tan θ)/(1 − cot θ) + (cot θ)/(1 − tan θ) = 1 + sec θ cosec θ.
Show answer
Let t = tan θ. LHS = t/(1 − 1/t) + (1/t)/(1 − t) = t²/(t − 1) − 1/(t(t − 1)) = (t³ − 1)/(t(t − 1)) = (t² + t + 1)/t = t + 1 + 1/t. Now t + 1/t = (sin²θ + cos²θ)/(sin θ cos θ) = sec θ cosec θ. So LHS = 1 + sec θ cosec θ. ∎
Answer: Proved
12. Quick Revision Summary
- sin = opp/hyp, cos = adj/hyp, tan = opp/adj; reciprocals cosec, sec, cot.
- Ratios depend only on the angle (similar triangles).
- Table: sin 0°…90° = √0/2 … √4/2; cos reversed; tan = sin/cos.
- sin²A + cos²A = 1; sec²A − tan²A = 1; cosec²A − cot²A = 1.
- Prove identities by converting to sin/cos or using these identities.