Class 10 Maths Chapter 8 Introduction to Trigonometry Practice Paper 4

Class: X Subject: Mathematics Session: 2025-26 Chapter: 08 - Introduction to Trigonometry Time: 1½ Hours Max. Marks: 40

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five sections A, B, C, D and E.

3. Section A contains 10 MCQs of 1 mark each.

4. Section B contains 4 questions of 2 marks each.

5. Section C contains 3 questions of 3 marks each.

6. Section D contains 1 question of 5 marks.

7. Section E contains 2 Case Study Based questions of 4 marks each.

SECTION A - Multiple Choice Questions (1 mark each)

1. If tan θ = 4/3, then the value of (3 sin θ + 2 cos θ)/(3 sin θ - 2 cos θ) is:
(a) 2
(b) 3
(c) 4
(d) 5

2. The value of (1 - cos²θ)(1 + cot²θ) is:
(a) 0
(b) 1
(c) sin²θ
(d) cos²θ

3. If sin(A - B) = √3/2 and cos(A + B) = 0, where A > B and A + B ≤ 90°, then the value of A is:
(a) 30°
(b) 45°
(c) 60°
(d) 75°

4. The value of (cos²0° + cos²45° + cos²60° + cos²90°) is:
(a) 1
(b) 7/4
(c) 9/4
(d) 2

5. If 3 tan²θ = 1, then the value of (2 sin²θ + 3 cos²θ) is:
(a) 7/3
(b) 8/3
(c) 10/3
(d) 11/3

6. If cosec A = √10, then the value of (sin A + cos A) is:
(a) √10
(b) (1 + 3)/√10
(c) 4/√10
(d) 1

7. The value of (tan 15° + cot 75°) is:
(a) 1
(b) 2 tan 15°
(c) 2 cot 75°
(d) 0

8. If sin A = 12/13, then the value of (13 sec A - 12 tan A) is:
(a) 1
(b) 5
(c) 12
(d) 13

9. Assertion
(a) : If 3 tan θ = 4, then (sin θ + cos θ) = 7/5 Reason (R): If 3 tan θ = 4, then sin θ = 4/5 and cos θ = 3/5
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

10. Assertion
(a) : The value of sin 60° cos 30° + sin 30° cos 60° is equal to 1 Reason (R): sin(A + B) = sin A cos B + cos A sin B
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

SECTION B - Short Answer Questions (2 marks each)

11. If cot A = 12/5, find the value of (2 sin A + 3 cos A)/(4 sin A + 3 cos A).

12. Evaluate: 2 tan²45° + cos²30° - sin²60°

13. If 3 sin²θ = 2 cos²θ, find the value of cot θ.

14. Prove that: (sin θ + cos θ)² + (sin θ - cos θ)² = 2

SECTION C - Short Answer Questions (3 marks each)

15. Prove that: (tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A

16. Prove that: (sin θ + 1 - cos θ)/(cos θ - 1 + sin θ) = (1 + sin θ)/cos θ OR If cosec θ - sin θ = a³ and sec θ - cos θ = b³, prove that a²b²(a² + b²) = 1

17. Prove that: (cos θ/(1 - tan θ)) + (sin θ/(1 - cot θ)) = sin θ + cos θ OR Prove that: tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + sec θ cosec θ

SECTION D - Long Answer Question (5 marks)

18.
(a) Prove that: (1 + sin A)/(1 - sin A) = (sec A + tan A)² [3]
(b) If a cos θ + b sin θ = m and a sin θ - b cos θ = n, prove that a² + b² = m² + n² [2]

SECTION E - Case Study Based Questions (4 marks each)

19. Leaning Tower Problem A boy standing at the base of a leaning tower observes that at a distance of 10 meters from the tower's base, the angle of elevation to the top is 60°. The tower leans at an angle such that it makes 75° with the ground. He wants to calculate the actual length of the tower and its vertical height. Based on the given information, answer the following questions:
(a) Find the vertical height of the point directly above the 10-meter mark [1]
(b) Find the value of tan 60° + cot 30° [1]
(c) If the tower were vertical, what would be its height for the same angle of elevation? [1]
(d) Verify that tan²60° - sec²60° = -1 [1]

20. Satellite Dish Installation A satellite dish needs to be installed on the roof of a building. The technician uses a ladder that is 20 meters long. The ladder is placed such that it makes an angle of 60° with the ground. After installing, he adjusts it to make an angle of 45° with the ground without changing the point where the ladder touches the ground. Based on the given information, answer the following questions:
(a) Find the height reached by the ladder at 60° angle (Take √3 = 1.732) [1]
(b) Find the height reached by the ladder at 45° angle (Take √2 = 1.414) [1]
(c) Find the difference in heights [1]
(d) Find sin²60° + cos²60° [1] DETAILED ANSWER KEY - PAPER 04

SECTION A - Answers to MCQs

1.
(b) 3

Solution:

Given: tan θ = 4/3 Let sin θ = 4/5 and cos θ = 3/5 (3 sin θ + 2 cos θ)/(3 sin θ - 2 cos θ) = (3×4/5 + 2×3/5)/(3×4/5 - 2×3/5) = (12/5 + 6/5)/(12/5 - 6/5) = (18/5)/(6/5) = 18/6 = 3

2.
(b) 1

Solution:

(1 - cos²θ)(1 + cot²θ) = sin²θ × cosec²θ = sin²θ × (1/sin²θ) = 1

3.
(d) 75°

Solution:

sin(A - B) = √3/2 → A - B = 60° cos(A + B) = 0 → A + B = 90° Adding: 2A = 150° → A = 75°

4.
(c) 9/4

Solution:

cos²0° + cos²45° + cos²60° + cos²90° = (1)² + (1/√2)² + (1/2)² + (0)² = 1 + 1/2 + 1/4 + 0 = (4 + 2 + 1)/4 = 7/4

5.
(c) 10/3

Solution:

3 tan²θ = 1 → tan²θ = 1/3 sec²θ = 1 + tan²θ = 1 + 1/3 = 4/3 cos²θ = 3/4 sin²θ = 1 - 3/4 = 1/4 2 sin²θ + 3 cos²θ = 2(1/4) + 3(3/4) = 1/2 + 9/4 = 2/4 + 9/4 = 11/4

6.
(c) 4/√10

Solution:

cosec A = √10 → sin A = 1/√10 cos A = √(1 - 1/10) = √(9/10) = 3/√10 sin A + cos A = 1/√10 + 3/√10 = 4/√10

7.
(b) 2 tan 15°

Solution:

tan 15° + cot 75° Note: cot 75° = cot(90° - 15°) = tan 15° = tan 15° + tan 15° = 2 tan 15°

8.
(b) 5

Solution:

Given: sin A = 12/13 cos A = 5/13, sec A = 13/5, tan A = 12/5 13 sec A - 12 tan A = 13(13/5) - 12(12/5) = 169/5 - 144/5 = 25/5 = 5

9.
(a)

Solution:

3 tan θ = 4 → tan θ = 4/3 If tan θ = 4/3, then sin θ = 4/5, cos θ = 3/5 sin θ + cos θ = 4/5 + 3/5 = 7/5 ✓ Both A and R are true, and R explains A. Answer:
(a)

10.
(a)

Solution:

sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1 ✓ This is actually the expansion of sin(60° + 30°) = sin 90° = 1 R explains why the sum equals 1. Answer:
(a)

SECTION B - Answers to Short Answer Questions

11.

Solution:

Given: cot A = 12/5, so tan A = 5/12 Let sin A = 5/13 and cos A = 12/13 (2 sin A + 3 cos A)/(4 sin A + 3 cos A) = (2×5/13 + 3×12/13)/(4×5/13 + 3×12/13) = (10/13 + 36/13)/(20/13 + 36/13) = (46/13)/(56/13) = 46/56 = 23/28 12.

Solution:

2 tan²45° + cos²30° - sin²60° = 2(1)² + (√3/2)² - (√3/2)² = 2(1) + 3/4 - 3/4 = 2 13.

Solution:

3 sin²θ = 2 cos²θ 3 sin²θ = 2(1 - sin²θ) 3 sin²θ = 2 - 2 sin²θ 5 sin²θ = 2 sin²θ = 2/5 cos²θ = 3/5 cot²θ = cos²θ/sin²θ = (3/5)/(2/5) = 3/2 cot θ = √(3/2) = √3/√2 = √(3/2) 14.

Solution:

(sin θ + cos θ)² + (sin θ - cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ + sin²θ - 2 sin θ cos θ + cos²θ = 2 sin²θ + 2 cos²θ = 2(sin²θ + cos²θ) = 2(1) = 2

SECTION C - Answers to Short Answer Questions

15.

Solution:

LHS = (tan A + sec A - 1)/(tan A - sec A + 1) We know: sec²A - tan²A = 1 (sec A + tan A)(sec A - tan A) = 1 Let sec A + tan A = p, then sec A - tan A = 1/p LHS = (p - 1)/(1/p + 1) = (p - 1)/[(1 + p)/p] = p(p - 1)/(1 + p) Now, p = sec A + tan A = (1 + sin A)/cos A After substitution and simplification: = (1 + sin A)/cos A = RHS 16.

Solution (Option 1):

LHS = (sin θ + 1 - cos θ)/(cos θ - 1 + sin θ) = (sin θ + 1 - cos θ)/(sin θ + cos θ - 1) Multiply numerator and denominator by (sin θ + cos θ + 1): After simplification using standard identities: = (1 + sin θ)/cos θ = RHS

Solution (Option 2 - OR):

Given: cosec θ - sin θ = a³ and sec θ - cos θ = b³ cosec θ - sin θ = 1/sin θ - sin θ = (1 - sin²θ)/sin θ = cos²θ/sin θ sec θ - cos θ = 1/cos θ - cos θ = (1 - cos²θ)/cos θ = sin²θ/cos θ a³ = cos²θ/sin θ and b³ = sin²θ/cos θ a²b² = (cos⁴θ/sin²θ)(sin⁴θ/cos²θ) = cos²θ sin²θ a² + b² = (cos⁴θ/sin²θ) + (sin⁴θ/cos²θ) After simplification: a²b²(a² + b²) = 1 17.

Solution (Option 1):

LHS = (cos θ/(1 - tan θ)) + (sin θ/(1 - cot θ)) = (cos θ/(1 - sin θ/cos θ)) + (sin θ/(1 - cos θ/sin θ)) = (cos θ/[(cos θ - sin θ)/cos θ]) + (sin θ/[(sin θ - cos θ)/sin θ]) = cos²θ/(cos θ - sin θ) + sin²θ/(sin θ - cos θ) = cos²θ/(cos θ - sin θ) - sin²θ/(cos θ - sin θ) = (cos²θ - sin²θ)/(cos θ - sin θ) = (cos θ + sin θ)(cos θ - sin θ)/(cos θ - sin θ) = cos θ + sin θ = RHS

Solution (Option 2 - OR):

LHS = tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = tan θ/(1 - 1/tan θ) + cot θ/(1 - tan θ) = tan θ/[(tan θ - 1)/tan θ] + cot θ/(1 - tan θ) = tan²θ/(tan θ - 1) - cot θ/(tan θ - 1) = (tan²θ - cot θ)/(tan θ - 1) After simplification: = 1 + sec θ cosec θ = RHS

SECTION D - Answer to Long Answer Question

18.
(a) Solution: LHS = (1 + sin A)/(1 - sin A) Multiply numerator and denominator 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]² = (1/cos A + sin A/cos A)² = (sec A + tan A)² = RHS
(b) Solution: Given: a cos θ + b sin θ = m ... (1) a sin θ - b cos θ = n ... (2) Squaring (1): a²cos²θ + 2ab sin θ cos θ + b²sin²θ = m² Squaring (2): a²sin²θ - 2ab sin θ cos θ + b²cos²θ = n² Adding: a²cos²θ + b²sin²θ + a²sin²θ + b²cos²θ = m² + n² a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ) = m² + n² a²(1) + b²(1) = m² + n² a² + b² = m² + n²

SECTION E - Answers to Case Study Based Questions

19.
(a) Vertical height: tan 60° = height/10 √3 = height/10 height = 10√3 = 10 × 1.732 = 17.32 meters
(b) tan 60° + cot 30°: = √3 + √3 = 2√3
(c) Height if tower were vertical: Same as part
(a) : 10√3 = 17.32 meters (The vertical height remains the same)
(d) Verification: tan²60° - sec²60° = (√3)² - (2)² = 3 - 4 = -1 ✓ 20.
(a) Height at 60° angle: sin 60° = height/20 √3/2 = height/20 height = 10√3 = 10 × 1.732 = 17.32 meters
(b) Height at 45° angle: sin 45° = height/20 1/√2 = height/20 height = 20/√2 = 20√2/2 = 10√2 = 10 × 1.414 = 14.14 meters
(c) Difference in heights:

Difference = 17.32 - 14.14 = 3.18 meters
(d) sin²60° + cos²60°: = (√3/2)² + (1/2)² = 3/4 + 1/4 = 1