NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry (NCERT 2026–27)

These Class 10 Maths Chapter 8 solutions cover Introduction to Trigonometry from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 8.1, Exercise 8.2 and Exercise 8.3 is reproduced exactly as in the book and solved step by step — finding trigonometric ratios, evaluating ratios of 30°, 45° and 60°, and proving trigonometric identities — so you can revise the chapter quickly and score full marks.

Class: 10 Subject: Mathematics Chapter: 8 Name: Introduction to Trigonometry Exercises: 8.1, 8.2, 8.3 Session: 2026–27
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Chapter 8 Overview

Chapter 8 of Class 10 Maths, Introduction to Trigonometry, studies the relationships between the sides and angles of a right triangle. You define the six trigonometric ratios — sine, cosine, tangent, cosecant, secant and cotangent — of an acute angle, learn that these ratios depend only on the angle (not on the size of the triangle), and compute them for the specific angles 0°, 30°, 45°, 60° and 90°. The chapter then establishes the three fundamental trigonometric identities — sin2A + cos2A = 1, 1 + tan2A = sec2A and 1 + cot2A = cosec2A — and uses them to express one ratio in terms of another and to prove a wide range of identities. The solutions below work through every question of Exercise 8.1, 8.2 and 8.3.

Key Concepts & Definitions

Trigonometric ratios: in a right triangle ABC right-angled at B, for the acute angle A — sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse, tan A = opposite/adjacent.

Reciprocal ratios: cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A.

Quotient relations: tan A = sin A / cos A and cot A = cos A / sin A.

Independence from size: the value of any trigonometric ratio of a given angle does not change with the lengths of the sides of the triangle — it depends only on the angle.

Bounds: sin A and cos A never exceed 1; sec A (0° ≤ A < 90°) and cosec A (0° < A ≤ 90°) are always ≥ 1.

Identity: an equation involving trigonometric ratios that is true for all admissible values of the angle.

Important Formulas (Chapter 8)

Pythagoras-based identities: sin2A + cos2A = 1;   1 + tan2A = sec2A;   1 + cot2A = cosec2A.

Reciprocal & quotient: cosec A = 1/sin A; sec A = 1/cos A; cot A = 1/tan A; tan A = sin A/cos A; cot A = cos A/sin A.

Standard values:

∠A0°30°45°60°90°
sin A01/21/√2√3/21
cos A1√3/21/√21/20
tan A01/√31√3Not defined
cosec ANot defined2√22/√31
sec A12/√3√22Not defined
cot ANot defined√311/√30

Exercise 8.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. In ▵ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos C

SOLUTION By Pythagoras theorem, AC2 = AB2 + BC2 = 242 + 72 = 576 + 49 = 625, so AC = 25 cm. For angle A: opposite side = BC = 7, adjacent = AB = 24, hypotenuse = AC = 25. (i) sin A = BC/AC = 7/25;   cos A = AB/AC = 24/25. For angle C: opposite side = AB = 24, adjacent = BC = 7. (ii) sin C = AB/AC = 24/25;   cos C = BC/AC = 7/25.

2. In Fig. 8.13, find tan P − cot R.

SOLUTION In the figure, ▵PQR is right-angled at Q with PR = 13 cm (hypotenuse) and PQ = 12 cm. QR = √(PR2 − PQ2) = √(132 − 122) = √(169 − 144) = √25 = 5 cm. tan P = QR/PQ = 5/12.   cot R = QR/PQ = 5/12 (for angle R, opposite = PQ, adjacent = QR, so cot R = adjacent/opposite = QR/PQ = 5/12). ∴ tan P − cot R = 5/12 − 5/12 = 0.

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

SOLUTION sin A = opposite/hypotenuse = 3/4, so in a right triangle take opposite = 3k, hypotenuse = 4k. Adjacent = √((4k)2 − (3k)2) = √(16k2 − 9k2) = √(7k2) = √7 k. cos A = adjacent/hypotenuse = √7 k / 4k = √7/4. tan A = sin A/cos A = (3/4) ÷ (√7/4) = 3/√7.

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

SOLUTION 15 cot A = 8 ⇒ cot A = 8/15 = adjacent/opposite, so take adjacent = 8k, opposite = 15k. Hypotenuse = √((8k)2 + (15k)2) = √(64k2 + 225k2) = √(289k2) = 17k. sin A = opposite/hypotenuse = 15k/17k = 15/17. sec A = hypotenuse/adjacent = 17k/8k = 17/8.

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

SOLUTION sec θ = hypotenuse/adjacent = 13/12, so take hypotenuse = 13k, adjacent = 12k. Opposite = √((13k)2 − (12k)2) = √(169k2 − 144k2) = √(25k2) = 5k. sin θ = 5/13;   cos θ = 12/13;   tan θ = 5/12. cosec θ = 13/5;   cot θ = 12/5.

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

SOLUTION Take a right triangle ABC right-angled at C, where ∠A and ∠B are the acute angles. cos A = AC/AB and cos B = BC/AB. Given cos A = cos B, so AC/AB = BC/AB, which gives AC = BC. In a triangle, sides opposite equal sides are equal, but here AC = BC means the angles opposite them are equal: ∠B = ∠A (angles opposite equal sides of triangle ABC). ∴ ∠A = ∠B.

7. If cot θ = 7/8, evaluate : (i) (1 + sin θ)(1 − sin θ) / [(1 + cos θ)(1 − cos θ)] (ii) cot2θ

SOLUTION cot θ = 7/8 = adjacent/opposite, so adjacent = 7k, opposite = 8k, hypotenuse = √(49k2 + 64k2) = √113 k. (i) (1 + sin θ)(1 − sin θ) = 1 − sin2θ = cos2θ, and (1 + cos θ)(1 − cos θ) = 1 − cos2θ = sin2θ. So the expression = cos2θ/sin2θ = cot2θ = (7/8)2 = 49/64. (ii) cot2θ = (7/8)2 = 49/64.

8. If 3 cot A = 4, check whether (1 − tan2A)/(1 + tan2A) = cos2A − sin2A or not.

SOLUTION 3 cot A = 4 ⇒ cot A = 4/3, so tan A = 3/4. Take opposite = 3k, adjacent = 4k, hypotenuse = √(9k2 + 16k2) = 5k. sin A = 3/5, cos A = 4/5. LHS = (1 − tan2A)/(1 + tan2A) = (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25. RHS = cos2A − sin2A = (4/5)2 − (3/5)2 = 16/25 − 9/25 = 7/25. LHS = RHS = 7/25, so the relation holds true.

9. In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C − sin A sin C

SOLUTION tan A = 1/√3 means ∠A = 30°, and since ∠B = 90°, ∠C = 60°. So sin A = 1/2, cos A = √3/2, sin C = √3/2, cos C = 1/2. (i) sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1. (ii) cos A cos C − sin A sin C = (√3/2)(1/2) − (1/2)(√3/2) = √3/4 − √3/4 = 0.

10. In ▵PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

SOLUTION Let QR = x, then PR = 25 − x. By Pythagoras theorem, PR2 = PQ2 + QR2. (25 − x)2 = 52 + x2 ⇒ 625 − 50x + x2 = 25 + x2 ⇒ 625 − 50x = 25 ⇒ 50x = 600 ⇒ x = 12. So QR = 12 cm and PR = 25 − 12 = 13 cm. For angle P: opposite = QR = 12, adjacent = PQ = 5, hypotenuse = PR = 13. sin P = QR/PR = 12/13;   cos P = PQ/PR = 5/13;   tan P = QR/PQ = 12/5.

11. State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1. (ii) sec A = 12/5 for some value of angle A. (iii) cos A is the abbreviation used for the cosecant of angle A. (iv) cot A is the product of cot and A. (v) sin θ = 4/3 for some angle θ.

SOLUTION (i) False. tan A can be greater than 1; for example tan 60° = √3 > 1. (ii) True. sec A = hypotenuse/adjacent ≥ 1, and 12/5 = 2.4 > 1, so it is a valid value of sec A for some angle. (iii) False. cos A is the abbreviation for the cosine of angle A, not the cosecant (which is cosec A). (iv) False. cot A means the cotangent of angle A; ‘cot’ separated from A has no meaning — it is not a product. (v) False. sin θ can never exceed 1, but 4/3 > 1, so no angle has sin θ = 4/3.

Exercise 8.2 Solutions

1. Evaluate the following : (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan2 45° + cos2 30° − sin2 60° (iii) cos 45° / (sec 30° + cosec 30°) (iv) (sin 30° + tan 45° − cosec 60°) / (sec 30° + cos 60° + cot 45°) (v) (5 cos2 60° + 4 sec2 30° − tan2 45°) / (sin2 30° + cos2 30°)

SOLUTION (i) = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. (ii) = 2(1)2 + (√3/2)2 − (√3/2)2 = 2 + 3/4 − 3/4 = 2. (iii) Numerator = cos 45° = 1/√2. Denominator = sec 30° + cosec 30° = 2/√3 + 2 = (2 + 2√3)/√3. So expression = (1/√2) ÷ [(2 + 2√3)/√3] = √3 / [√2 · 2(1 + √3)] = √3 / [2√2(1 + √3)]. Rationalise by multiplying by (√3 − 1) and by √2: = √3(√3 − 1)√2 / [2√2 · 2 · √2] … simplifying gives (3√2 − √6)/8. (iv) Numerator = 1/2 + 1 − 2/√3 = 3/2 − 2/√3. Denominator = 2/√3 + 1/2 + 1 = 2/√3 + 3/2. Take a common form: Numerator = (3√3 − 4)/(2√3), Denominator = (4 + 3√3)/(2√3). So expression = (3√3 − 4)/(3√3 + 4). Rationalising by (3√3 − 4): = (3√3 − 4)2/[(3√3)2 − 42] = (27 − 24√3 + 16)/(27 − 16) = (43 − 24√3)/11. (v) Numerator = 5(1/2)2 + 4(2/√3)2 − (1)2 = 5/4 + 16/3 − 1 = (15 + 64 − 12)/12 = 67/12. Denominator = sin230° + cos230° = 1. ∴ expression = 67/12.

2. Choose the correct option and justify your choice : (i) 2 tan 30° / (1 + tan2 30°)   (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30° (ii) (1 − tan2 45°) / (1 + tan2 45°)   (A) tan 90° (B) 1 (C) sin 45° (D) 0 (iii) sin 2A = 2 sin A is true when A =   (A) 0° (B) 30° (C) 45° (D) 60° (iv) 2 tan 30° / (1 − tan2 30°)   (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

SOLUTION (i) tan 30° = 1/√3, so 2(1/√3)/(1 + 1/3) = (2/√3)/(4/3) = (2/√3)(3/4) = 6/(4√3) = √3/2 = sin 60°. Answer (A) sin 60°. (ii) tan 45° = 1, so (1 − 1)/(1 + 1) = 0/2 = 0. Answer (D) 0. (iii) sin 2A = 2 sin A. At A = 0°: sin 0° = 0 and 2 sin 0° = 0, so it holds. (At 30°, 45°, 60° the two sides differ.) Answer (A) 0°. (iv) 2(1/√3)/(1 − 1/3) = (2/√3)/(2/3) = (2/√3)(3/2) = 3/√3 = √3 = tan 60°. Answer (C) tan 60°.

3. If tan (A + B) = √3 and tan (A − B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.

SOLUTION tan (A + B) = √3 = tan 60° ⇒ A + B = 60° …(1) tan (A − B) = 1/√3 = tan 30° ⇒ A − B = 30° …(2) Adding (1) and (2): 2A = 90° ⇒ A = 45°. Subtracting (2) from (1): 2B = 30° ⇒ B = 15°. ∴ A = 45° and B = 15°.

4. State whether the following are true or false. Justify your answer. (i) sin (A + B) = sin A + sin B. (ii) The value of sin θ increases as θ increases. (iii) The value of cos θ increases as θ increases. (iv) sin θ = cos θ for all values of θ. (v) cot A is not defined for A = 0°.

SOLUTION (i) False. Take A = 30°, B = 60°: sin (A + B) = sin 90° = 1, but sin 30° + sin 60° = 1/2 + √3/2 ≠ 1. (ii) True. As θ increases from 0° to 90°, sin θ increases from 0 to 1. (iii) False. As θ increases from 0° to 90°, cos θ decreases from 1 to 0. (iv) False. sin θ = cos θ only at θ = 45°, not for all θ. (v) True. cot 0° = cos 0°/sin 0° = 1/0, which is not defined.

Exercise 8.3 Solutions

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

SOLUTION Using cosec2A = 1 + cot2A: cosec A = √(1 + cot2A), so sin A = 1/cosec A = 1/√(1 + cot2A). cos A = cot A · sin A = cot A/√(1 + cot2A), so sec A = 1/cos A = √(1 + cot2A)/cot A. tan A = 1/cot A = 1/cot A.

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

SOLUTION cos A = 1/sec A. sin A = √(1 − cos2A) = √(1 − 1/sec2A) = √(sec2A − 1)/sec A. cosec A = 1/sin A = sec A/√(sec2A − 1). tan A = √(sec2A − 1) (since sec2A − 1 = tan2A). cot A = 1/tan A = 1/√(sec2A − 1).

3. Choose the correct option. Justify your choice. (i) 9 sec2A − 9 tan2A =   (A) 1 (B) 9 (C) 8 (D) 0 (ii) (1 + tan θ + sec θ)(1 + cot θ − cosec θ) =   (A) 0 (B) 1 (C) 2 (D) −1 (iii) (sec A + tan A)(1 − sin A) =   (A) sec A (B) sin A (C) cosec A (D) cos A (iv) (1 + tan2A)/(1 + cot2A) =   (A) sec2A (B) −1 (C) cot2A (D) tan2A

SOLUTION (i) 9(sec2A − tan2A) = 9 × 1 = 9. Answer (B) 9. (ii) Write in sin/cos: (1 + tan θ + sec θ)(1 + cot θ − cosec θ) expands to 2. (Multiply out: = 1 + cot θ − cosec θ + tan θ + 1 − sec θ + sec θ + cosec θ − 1, and tan θ cot θ = 1, sec θ cosec θ cancels the cross terms; simplification gives 2.) Answer (C) 2. (iii) (sec A + tan A)(1 − sin A) = (1/cos A + sin A/cos A)(1 − sin A) = [(1 + sin A)(1 − sin A)]/cos A = (1 − sin2A)/cos A = cos2A/cos A = cos A. Answer (D) cos A. (iv) (1 + tan2A)/(1 + cot2A) = sec2A/cosec2A = (1/cos2A)/(1/sin2A) = sin2A/cos2A = tan2A. Answer (D) tan2A.

4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) (cosec θ − cot θ)2 = (1 − cos θ)/(1 + cos θ) (ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A (iii) tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ (iv) (1 + sec A)/sec A = sin2A/(1 − cos A) (v) (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A, using the identity cosec2A = 1 + cot2A. (vi) √[(1 + sin A)/(1 − sin A)] = sec A + tan A (vii) (sin θ − 2 sin3θ)/(2 cos3θ − cos θ) = tan θ (viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A (ix) (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A) (x) (1 + tan2A)/(1 + cot2A) = [(1 − tan A)/(1 − cot A)]2 = tan2A

SOLUTION (i) LHS = (cosec θ − cot θ)2 = (1/sin θ − cos θ/sin θ)2 = [(1 − cos θ)/sin θ]2 = (1 − cos θ)2/sin2θ. Replace sin2θ = 1 − cos2θ = (1 − cos θ)(1 + cos θ): = (1 − cos θ)2/[(1 − cos θ)(1 + cos θ)] = (1 − cos θ)/(1 + cos θ) = RHS. ✓ (ii) LHS = [cos2A + (1 + sin A)2]/[(1 + sin A)cos A] = [cos2A + 1 + 2 sin A + sin2A]/[(1 + sin A)cos A]. Using cos2A + sin2A = 1: = [2 + 2 sin A]/[(1 + sin A)cos A] = 2(1 + sin A)/[(1 + sin A)cos A] = 2/cos A = 2 sec A = RHS. ✓ (iii) Write in sin/cos. tan θ/(1 − cot θ) = (sin θ/cos θ)/(1 − cos θ/sin θ) = sin2θ/[cos θ(sin θ − cos θ)]. Similarly cot θ/(1 − tan θ) = cos2θ/[sin θ(cos θ − sin θ)] = −cos2θ/[sin θ(sin θ − cos θ)]. Adding: = [sin3θ − cos3θ]/[sin θ cos θ(sin θ − cos θ)]. Factor sin3θ − cos3θ = (sin θ − cos θ)(sin2θ + sin θ cos θ + cos2θ) = (sin θ − cos θ)(1 + sin θ cos θ). So = (1 + sin θ cos θ)/(sin θ cos θ) = 1 + 1/(sin θ cos θ) = 1 + sec θ cosec θ = RHS. ✓ (iv) LHS = (1 + sec A)/sec A = 1/sec A + 1 = cos A + 1. RHS = sin2A/(1 − cos A) = (1 − cos2A)/(1 − cos A) = [(1 − cos A)(1 + cos A)]/(1 − cos A) = 1 + cos A. LHS = RHS. ✓ (v) Divide numerator and denominator of LHS by sin A and use cos A/sin A = cot A, 1/sin A = cosec A. LHS = (cot A − 1 + cosec A)/(cot A + 1 − cosec A). Multiply numerator and denominator suitably and use cosec2A − cot2A = 1 i.e. (cosec A − cot A)(cosec A + cot A) = 1, replacing the ‘−1’ in numerator by −(cosec2A − cot2A): numerator = (cot A + cosec A) − (cosec A − cot A)(cosec A + cot A) = (cot A + cosec A)[1 − (cosec A − cot A)]. This equals (cot A + cosec A)(cot A + 1 − cosec A) which is (cot A + cosec A) × denominator. So LHS = cot A + cosec A = cosec A + cot A = RHS. ✓ (vi) Multiply inside the root by (1 + sin A): √[(1 + sin A)2/((1 − sin A)(1 + sin A))] = √[(1 + sin A)2/(1 − sin2A)] = √[(1 + sin A)2/cos2A] = (1 + sin A)/cos A = 1/cos A + sin A/cos A = sec A + tan A = RHS. ✓ (vii) LHS = sin θ(1 − 2 sin2θ)/[cos θ(2 cos2θ − 1)]. Now 1 − 2 sin2θ = 1 − 2(1 − cos2θ) = 2 cos2θ − 1. So the bracket factors cancel: LHS = sin θ/cos θ = tan θ = RHS. ✓ (viii) LHS = sin2A + 2 sin A cosec A + cosec2A + cos2A + 2 cos A sec A + sec2A. Since sin A cosec A = 1 and cos A sec A = 1, and sin2A + cos2A = 1: = 1 + 2 + cosec2A + 2 + sec2A = 5 + (1 + cot2A) + (1 + tan2A) = 7 + tan2A + cot2A = RHS. ✓ (ix) LHS = (cosec A − sin A)(sec A − cos A) = (1/sin A − sin A)(1/cos A − cos A) = [(1 − sin2A)/sin A][(1 − cos2A)/cos A] = (cos2A/sin A)(sin2A/cos A) = sin A cos A. RHS = 1/(tan A + cot A) = 1/(sin A/cos A + cos A/sin A) = 1/[(sin2A + cos2A)/(sin A cos A)] = sin A cos A. LHS = RHS. ✓ (x) First part: (1 + tan2A)/(1 + cot2A) = sec2A/cosec2A = sin2A/cos2A = tan2A. Second part: [(1 − tan A)/(1 − cot A)]2. Now 1 − cot A = 1 − 1/tan A = (tan A − 1)/tan A, so (1 − tan A)/(1 − cot A) = (1 − tan A) × tan A/(tan A − 1) = −tan A. Squaring gives tan2A. Hence both expressions equal tan2A. ✓

Common Mistakes to Avoid

Watch out for these

  • Writing sin A > 1 or cos A > 1 — these ratios never exceed 1, so values like sin θ = 4/3 are impossible.
  • Confusing which side is opposite and which is adjacent — it changes when you switch from angle A to angle C in the same triangle.
  • Treating sin (A + B) as sin A + sin B — trigonometric ratios do not distribute over addition.
  • Forgetting that sin2A means (sin A)2, not sin(A2).
  • Using cot 0° or tan 90° as if they were defined — both are not defined.
  • In identity proofs, not picking a single strategy (convert all to sin/cos, or simplify each side separately) — jumping between methods causes errors.

Practice MCQs & Assertion–Reason

1. If sin A = 3/5, then cos A equals:

(a) 4/5    (b) 5/4    (c) 3/4    (d) 5/3

2. The value of sin 60° cos 30° + sin 30° cos 60° is:

(a) 0    (b) 1    (c) 1/2    (d) √3/2

3. sec2A − tan2A is equal to:

(a) 0    (b) 1    (c) −1    (d) 2

4. The value of tan 60° is:

(a) 1/√3    (b) 1    (c) √3    (d) 2

5. If tan A = 1, then the value of 2 sin A cos A is:

(a) 0    (b) 1    (c) 1/2    (d) 2

6. If cos A = cos B (A, B acute), then:

(a) A > B    (b) A < B    (c) A = B    (d) A + B = 90°

7. (1 + tan2A) is equal to:

(a) cosec2A    (b) sec2A    (c) cot2A    (d) sin2A

8. If sec θ = 13/12, then sin θ equals:

(a) 12/13    (b) 5/13    (c) 5/12    (d) 13/5

9. The value of (1 − tan245°)/(1 + tan245°) is:

(a) tan 90°    (b) 1    (c) sin 45°    (d) 0

10. (sec A + tan A)(1 − sin A) is equal to:

(a) sec A    (b) sin A    (c) cos A    (d) cosec A

Answer key: 1-(a), 2-(b), 3-(b), 4-(c), 5-(b), 6-(c), 7-(b), 8-(b), 9-(d), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The value of sin θ can never be 4/3.

Reason: In a right triangle the opposite side can never be longer than the hypotenuse, so sin θ ≤ 1.

A-R 2. Assertion: sec2A − tan2A = 1 for all acute angles A.

Reason: Dividing sin2A + cos2A = 1 by cos2A gives 1 + tan2A = sec2A.

A-R 3. Assertion: tan 45° = 1.

Reason: In a right triangle with one 45° angle, the two legs are equal, so opposite = adjacent.

A-R 4. Assertion: cos θ increases as θ increases from 0° to 90°.

Reason: cos 0° = 1 and cos 90° = 0.

A-R 5. Assertion: If tan A = 1/√3, then A = 30°.

Reason: tan 30° = 1/√3 from the standard table of trigonometric ratios.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • sin A = opp/hyp, cos A = adj/hyp, tan A = opp/adj; cosec, sec, cot are their reciprocals.
  • tan A = sin A/cos A and cot A = cos A/sin A.
  • Trigonometric ratios of an angle depend only on the angle, not on the size of the triangle.
  • sin A and cos A are never greater than 1; sec A and cosec A are never less than 1.
  • Standard values for 0°, 30°, 45°, 60°, 90° should be memorised (see the table above).
  • Identities: sin2A + cos2A = 1;  1 + tan2A = sec2A;  1 + cot2A = cosec2A.
  • Use these identities to express any one ratio in terms of another and to prove other identities.

How to score full marks in this chapter

Memorise the standard-angle table cold — most of Exercise 8.2 is just substitution. For ratio questions, always sketch the right triangle and label opposite, adjacent and hypotenuse before writing ratios. For identity proofs, fix one strategy: either convert everything to sin and cos, or simplify each side separately; rationalise by multiplying by a conjugate (as in √[(1 + sin A)/(1 − sin A)]). Always state which identity you use, and keep each step on its own line so every step earns its mark.

Frequently Asked Questions

What is Class 10 Maths Chapter 8 about?

Chapter 8, Introduction to Trigonometry, defines the six trigonometric ratios (sin, cos, tan, cosec, sec, cot) of an acute angle in a right triangle, gives their values for 0°, 30°, 45°, 60° and 90°, and proves the fundamental trigonometric identities used to simplify and prove expressions.

How many exercises are there in Class 10 Maths Chapter 8?

There are three exercises — Exercise 8.1 (trigonometric ratios), Exercise 8.2 (ratios of specific angles) and Exercise 8.3 (trigonometric identities) — all solved step by step on this page.

What are the three main trigonometric identities in this chapter?

They are sin2A + cos2A = 1, 1 + tan2A = sec2A, and 1 + cot2A = cosec2A. They are derived from the Pythagoras theorem and hold for the angles where the ratios are defined.

Are these Class 10 Maths Chapter 8 solutions free?

Yes. All solutions are free and follow the official NCERT Class 10 Mathematics textbook for the 2026–27 session, with every answer verified against the book.

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