Chapter 8: Introduction to Trigonometry

Mathematics · CBSE Class 10 · NCERT Solutions

📖 Chapter 8 — Introduction to Trigonometry (CBSE Class 10 Maths): Detailed notes — trigonometric ratios, the standard-angle table and trigonometric identities.

Chapter 8 — Introduction to Trigonometry

📐 1. Trigonometric Ratios

In a right triangle (right angle not at the angle θ), for an acute angle θ:

sin θ = Opposite/Hypotenuse · cos θ = Adjacent/Hypotenuse · tan θ = Opposite/Adjacent

Reciprocals: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ, and tan θ = sin θ/cos θ.

📊 2. Standard Angle Values

θ	0°	30°	45°	60°	90°
sin	0	½	1/√2	√3/2	1
cos	1	√3/2	1/√2	½	0
tan	0	1/√3	1	√3	∞

🔗 3. Trigonometric Identities

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

↔️ 4. Complementary Angles

sin(90°−θ) = cos θ, cos(90°−θ) = sin θ, tan(90°−θ) = cot θ (and reciprocals).

🧠 Remember: sin=opp/hyp, cos=adj/hyp, tan=opp/adj. Learn the 0–90° table. Identities: sin²+cos²=1, 1+tan²=sec², 1+cot²=cosec². sin(90−θ)=cosθ. Build a right triangle from a given ratio to find the others.

⭐ Exam Tips: ① Find all ratios from one given ratio (make a right triangle). ② Evaluate expressions using the standard-angle table. ③ Prove identities (convert everything to sin & cos). ④ Use 9sec²A−9tan²A = 9. ⑤ Complementary-angle simplifications.

📘 EXERCISE 8.1 (Trigonometric Ratios)

1In △ABC right-angled at B, AB = 24 cm, BC = 7 cm. Find sin A, cos A, sin C, cos C.

right angle at B, AB = 24, BC = 7 Find hypotenuse AC by Pythagoras: AC = √(AB² + BC²) = √(576 + 49) = √625 = 25. For angle A: opposite = BC = 7, adjacent = AB = 24 → sin A = 7/25, cos A = 24/25. For angle C: opposite = AB = 24, adjacent = BC = 7 → sin C = 24/25, cos C = 7/25. ∴ sin A = 7/25, cos A = 24/25 ; sin C = 24/25, cos C = 7/25.

3If sin A = 3/4, find cos A and tan A.

Use sin²A + cos²A = 1 → cos A = √(1 − sin²A) = √(1 − 9/16) = √(7/16). cos A = √7/4. tan A = sin A / cos A = (3/4) ÷ (√7/4) = 3/√7. ∴ cos A = √7/4, tan A = 3/√7.

4Given 15 cot A = 8, find sin A and sec A.

cot A = 8/15, so in a right triangle adjacent = 8, opposite = 15. Hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = 17. sin A = opp/hyp = 15/17 ; sec A = hyp/adj = 17/8. ∴ sin A = 15/17, sec A = 17/8.

5Given sec θ = 13/12, find all other trigonometric ratios.

sec θ = hyp/adj = 13/12 → hyp = 13, adjacent = 12. Opposite = √(13² − 12²) = √(169 − 144) = √25 = 5. sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5. ∴ sin 5/13, cos 12/13, tan 5/12, cosec 13/5, cot 12/5.

📗 EXERCISE 8.2 (Standard Angles)

1Evaluate the expressions.

(i) sin60°cos30° + sin30°cos60° = (√3/2)(√3/2) + (½)(½) = 3/4 + 1/4 = 1. (ii) 2tan²45° + cos²30° − sin²60° = 2(1) + (√3/2)² − (√3/2)² = 2 + 0 = 2. (v) (5cos²60° + 4sec²30° − tan²45°)/(sin²30° + cos²30°) = (5·¼ + 4·(4/3) − 1)/1 = (5/4 + 16/3 − 1) = 67/12. ∴ (i) 1, (ii) 2, (v) 67/12 [ (iii) (3√2−√6)/8 ; (iv) (43−24√3)/11 ]

📙 EXERCISE 8.3 (Identities)

1Express sin A, sec A and tan A in terms of cot A.

Use 1 + cot²A = cosec²A → cosec A = √(1 + cot²A). sin A = 1/cosec A = 1/√(1 + cot²A). tan A = 1/cot A ; and sec A = √(1 + tan²A) = √(1 + 1/cot²A) = √(1 + cot²A)/cot A. ∴ sin A = 1/√(1+cot²A), tan A = 1/cot A, sec A = √(1+cot²A)/cot A.

3Evaluate 9 sec²A − 9 tan²A.

Take 9 common: 9(sec²A − tan²A). By the identity sec²A − tan²A = 1. = 9 × 1. ∴ 9 (option B)

Chapter 8: Introduction to Trigonometry