Class 10 Mathematics

Chapter 8 Solutions — Introduction to Trigonometry

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Solutions

Overview

Step-by-step NCERT solutions for Introduction to Trigonometry (Chapter 8, CBSE Class 10 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Introduction to Trigonometry textbook chapter.

Solved

What these solutions cover

All 19 questions in Introduction to Trigonometry are solved in the PDF. Here's what's inside, exercise by exercise:

Trigonometric Ratios

1.
In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
- (i) sin A, cos A
- (ii) sin C, cos C
In Fig. 8.13, find tan P – cot R. (Triangle PQR right-angled at Q, with PQ = 12 cm, QR = 5 cm, PR = 13 cm)
If sin A = 3/4, calculate cos A and tan A.
Given 15 cot A = 8, find sin A and sec A.
Given sec θ = 13/12, calculate all other trigonometric ratios.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
7.
If cot θ = 7/8, evaluate:
- (i) (1 + sin θ)(1 – sin θ) / [(1 + cos θ)(1 – cos θ)]
- (ii) cot² θ
If 3 cot A = 4, check whether (1 – tan²A) / (1 + tan²A) = cos²A – sin²A or not.
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
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.
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.

Trigonometric Ratios of Specific Angles

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

Trigonometric Identities

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Write all the other trigonometric ratios of angle A in terms of sec A.
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
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 [Hint: Write the expression in terms of sin θ and cos θ]
- (iv) (1 + sec A) / sec A = sin²A / (1 – cos A)…

Sample

Worked examples

3 questions from Introduction to Trigonometry, solved step by step.

Trigonometric Ratios

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

Using the Pythagorean theorem: AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625

⇒ AC = 25 cm

(i) For angle A: opposite side = BC = 7, adjacent = AB = 24, hypotenuse = AC = 25

sin A = BC/AC = 7/25

cos A = AB/AC = 24/25

(ii) For angle C: opposite side = AB = 24, adjacent = BC = 7, hypotenuse = AC = 25

sin C = AB/AC = 24/25

cos C = BC/AC = 7/25

Answer: sin A = 7/25, cos A = 24/25; sin C = 24/25, cos C = 7/25

Trigonometric Ratios of Specific Angles

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

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°

Verification: A + B = 60° ≤ 90° ✓, A > B (45° > 15°) ✓

Answer: A = 45°, B = 15°

Trigonometric Identities

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

Using the identity: cosec²A = 1 + cot²A

⇒ 1/sin²A = 1 + cot²A

⇒ sin²A = 1 / (1 + cot²A)

⇒ sin A = 1 / √(1 + cot²A) (positive since A is acute)

Since cos A = sin A · cot A:

cos A = cot A / √(1 + cot²A)

sec A = 1/cos A = √(1 + cot²A) / cot A

tan A = 1/cot A

Summary:

sin A = 1/√(1 + cot²A)

sec A = √(1 + cot²A) / cot A

tan A = 1/cot A

Answer: sin A = 1/√(1 + cot²A), sec A = √(1 + cot²A)/cot A, tan A = 1/cot A

All 19 questions in Introduction to Trigonometry are solved step by step in the Solutions PDF above.

Board papers

Introduction to Trigonometry in CBSE board papers

Tested in 115 questions across 23 Class 10 board papers (2020–2026). Open a set to see exactly how this chapter is asked:

Keep solving

Chapter 8 Solutions — Introduction to Trigonometry

Overview

What these solutions cover

Trigonometric Ratios

Trigonometric Ratios of Specific Angles

Trigonometric Identities

Worked examples

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

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

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

Introduction to Trigonometry in CBSE board papers

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