Class 10 Maths Chapter 8 Introduction to Trigonometry PYQ

Amitesh Nagar, Indore (M.P.) Class: X Subject: Mathematics Session: 2025-26 Chapter: Ch 8: Introduction to Trigonometry (PYQ) PREVIOUS YEAR QUESTIONS (PYQ) Chapter 8: Introduction to Trigonometry CBSE Board Exam 2019–2025 | With Direct Answers This document contains chapter-wise Previous Year Questions from CBSE Class X Board Examinations (2019–2025) for Chapter 8: Introduction to Trigonometry . Each question includes the year of examination, marks allotted, and direct answer for quick revision. ⚠ NOTE: All questions strictly as per CBSE 2025–26 Syllabus. Topics: Trigonometric Ratios of acute angles, Standard angle values (0°, 30°, 45°, 60°, 90°), Trigonometric Identities. EXCLUDED:

Complementary angle ratios (deleted from syllabus).

SECTION A: Multiple Choice Questions (1 Mark Each)

[CBSE 2024 | 1 Mark]

Q1. If sin α = √3/2 and cos β = √3/2, then tan α · tan β is:
(a) √3
(b) 1/√3
(c) 1
(d) 0 Ans:
(c) 1. sin α = √3/2 ⇒ α = 60°. cos β = √3/2 ⇒ β = 30°. tan 60° × tan 30° = √3 × 1/√3 = 1 [CBSE 2023 | 1 Mark]

Q2. If sin θ = cos θ, then the value of θ is:
(a) 0°
(b) 30°
(c) 45°
(d) 60° Ans:
(c) 45°. sin θ = cos θ ⇒ tan θ = 1 ⇒ θ = 45° [CBSE 2024 | 1 Mark]

Q3. If tan θ = 1/√3, then the value of sec²θ − cosec²θ is:
(a) −8/3
(b) 8/3
(c) −4/3
(d) 4/3 Ans:
(a) −8/3. tan θ = 1/√3 ⇒ θ = 30°. sec²30° − cosec²30° = 4/3 − 4 = −8/3 Amitesh Nagar, Indore (M.P.) [CBSE 2021 | 1 Mark]

Q4. The value of (sin 30° + cos 60°) − (sin 60° + cos 30°) is:
(a) 0
(b) 1 + √3
(c) 1 − √3
(d) 1 + 2√3 Ans:
(c) 1 − √3. (1/2 + 1/2) − (√3/2 + √3/2) = 1 − √3 [CBSE 2020 | 1 Mark]

Q5. The value of sin²60° + cos²60° is:
(a) 0
(b) 1
(c) 1/2
(d) √3/2 Ans:
(b) 1. By identity sin²θ + cos²θ = 1 for all θ. [CBSE 2021 | 1 Mark]

Q6. Given that sin θ = a/b, then cos θ is:
(a) b/√(b² − a²)
(b) b/a
(c) √(b² − a²)/b
(d) a/√(b² − a²) Ans:
(c) √(b² − a²)/b. cos θ = √(1 − a²/b²) = √(b² − a²)/b [CBSE 2023 | 1 Mark]

Q7. If sin θ − cos θ = 0, then the value of sin⁴θ + cos⁴θ is:
(a) 1
(b) 3/4
(c) 1/2
(d) 1/4 Ans:
(c) 1/2. sin θ = cos θ ⇒ θ = 45°. sin⁴45° + cos⁴45° = 1/4 + 1/4 = 1/2 [CBSE 2019 | 1 Mark]

Q8. If sec θ = 25/7, then sin θ =
(a) 7/25
(b) 24/25
(c) 25/24
(d) 7/24 Ans:
(b) 24/25. cos θ = 7/25. sin θ = √(1 − 49/625) = √(576/625) = 24/25 [CBSE 2020 | 1 Mark]

Q9. If 4 tan θ = 3, then the value of (4 sin θ − cos θ)/(4 sin θ + cos θ) is:
(a) 2/3
(b) 1/3
(c) 1/2
(d) 3/4 Ans:
(c) 1/2. tan θ = 3/4. Divide by cos θ: (4 tan θ − 1)/(4 tan θ + 1) = (3 − 1)/(3 + 1) = Amitesh Nagar, Indore (M.P.) [CBSE 2022 | 1 Mark]

Q10. The maximum value of 1/sec θ is:
(a) 0
(b) 1
(c) √2
(d) 1/2 Ans:
(b) 1. 1/sec θ = cos θ. Maximum value of cos θ = 1 (at θ = 0°). [CBSE 2022 | 1 Mark]

Q11. If x = a cos θ and y = b sin θ, then b²x² + a²y² =
(a) a²b²
(b) ab
(c) a⁴b⁴
(d) a² + b² Ans:
(a) a²b². b²(a cos θ)² + a²(b sin θ)² = a²b²(cos²θ + sin²θ) = a²b²

SECTION B: Assertion-Reason Questions (1 Mark Each)

[CBSE 2023 | 1 Mark]

Q12. Assertion
(a) : For any acute angle θ, sin²θ + cos²θ = 1. Reason (R): For any acute angle θ, sec²θ − tan²θ = 1.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true Ans:
(b) Both true but R is not the correct explanation of A. Both are independent identities. [CBSE 2024 | 1 Mark]

Q13. Assertion
(a) : In a right ΔABC, right-angled at B, cosec A can be less than 1. Reason (R): cosec A = Hypotenuse/Perpendicular, and hypotenuse is always the longest side.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true Ans:
(d) A is false but R is true. Since hypotenuse ≥ perpendicular, cosec A ≥ 1 always.

SECTION C: Short Answer Questions (2 Marks Each)

[CBSE 2024 | 2 Marks]

Q14. If tan A + cot A = 6, find the value of tan²A + cot²A − 4. Ans: (tan A + cot A)² = 36 ⇒ tan²A + cot²A + 2 = 36 ⇒ tan²A + cot²A = 34. Answer = 34 − 4 = 30 Amitesh Nagar, Indore (M.P.) [CBSE 2020 | 2 Marks]

Q15. Evaluate: 2 tan²45° + cos²30° − sin²60°. Ans: 2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2 [CBSE 2019 | 2 Marks]

Q16. In a right triangle ABC, right-angled at B, if tan A = 1, verify that 2 sin A cos A = 1. Ans: tan A = 1 ⇒ A = 45°. 2 sin 45° cos 45° = 2 × (1/√2)(1/√2) = 2 × 1/2 = 1. Verified. [CBSE 2022 | 2 Marks]

Q17. If sin(A − B) = 1/2 and cos(A + B) = 1/2, where 0° < A + B ≤ 90°, find A and B. Ans: A − B = 30° ... (i). A + B = 60° ... (ii). Adding: 2A = 90° ⇒ A = 45°, B = 15°. [CBSE 2021 | 2 Marks]

Q18. If √3 tan θ = 1, find the value of sin²θ − cos²θ. Ans: tan θ = 1/√3 ⇒ θ = 30°. sin²30° − cos²30° = 1/4 − 3/4 = −1/2

SECTION D: Prove the Following Identities (3 Marks Each)

[CBSE 2024 | 3 Marks]

Q19. Prove that: (cosec θ − sin θ)(sec θ − cos θ)(tan θ + cot θ) = 1 Ans: LHS = [cos²θ/sin θ] × [sin²θ/cos θ] × [1/(sin θ cos θ)] = (sin θ cos θ)/(sin θ cos θ) = 1 = RHS [CBSE 2023 | 3 Marks]

Q20. Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ Ans: LHS = sin²θ + cosec²θ + 2 + cos²θ + sec²θ + 2 = 1 + (1 + cot²θ) + (1 + tan²θ) + 4 = 7 + tan²θ + cot²θ = RHS [CBSE 2022 | 3 Marks]

Q21. Prove that: tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ Ans: LHS = tan²θ/(tan θ − 1) − 1/[tan θ(tan θ − 1)] = (tan³θ − 1)/[tan θ(tan θ − 1)] = (tan²θ + tan θ + 1)/tan θ = tan θ + 1 + cot θ = 1 + 1/(sin θ cos θ) = 1 + sec θ cosec θ = RHS [CBSE 2021 | 3 Marks]

Q22. Prove that: (sin A − 2 sin³A)/(2 cos³A − cos A) = tan A Ans: LHS = sin A(1 − 2 sin²A) / cos A(2 cos²A − 1). Since 2 cos²A − 1 = 1 − 2 sin²A, LHS = sin A/cos A = tan A = RHS [CBSE 2020 | 3 Marks]

Q23. Prove that: √[(1 + sin A)/(1 − sin A)] = sec A + tan A Ans: Multiply inside by (1 + sin A)/(1 + sin A): √[(1 + sin A)²/(1 − sin²A)] = (1 + sin A)/cos A = sec A + tan A = RHS Amitesh Nagar, Indore (M.P.) [CBSE 2019 | 3 Marks]

Q24. Prove that: (cosec A − cot A)² = (1 − cos A)/(1 + cos A) Ans: LHS = [(1 − cos A)/sin A]² = (1 − cos A)²/sin²A = (1 − cos A)²/(1 − cos²A) = (1 − cos A)/(1 + cos A) = RHS

SECTION E: Long Answer Questions (5 Marks Each)

[CBSE 2022 | 5 Marks]

Q25. If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1. Ans: Squaring: 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 [CBSE 2019 | 5 Marks]

Q26. If tan θ + sin θ = m and tan θ − sin θ = n, show that m² − n² = 4√(mn). Ans: m² − n² = 4 tan θ sin θ. mn = tan²θ − sin²θ = tan²θ sin²θ. √(mn) = tan θ sin θ. So 4√(mn) = 4 tan θ sin θ = m² − n²

SECTION F: Case Study Based Questions (4 Marks Each)

[CBSE 2025 | 4 Marks]

Q27. Case Study: In a right triangle ABC, right-angled at B, AB = 5 cm and AC = 13 cm. (i) Find sin A and cos A. (ii) Find tan A and cot A. (iii) Verify that sin²A + cos²A = 1. (iv) Find sec²A − tan²A. Ans: BC = √(169 − 25) = 12 cm. (i) sin A = 12/13, cos A = 5/13. (ii) tan A = 12/5, cot A = 5/12. (iii) 144/169 + 25/169 = 1 ✓ (iv) sec²A − tan²A = 1 (identity) [CBSE 2024 | 4 Marks]

Q28. Case Study: For an acute angle θ in a right triangle with sides p, b and h: (i) If sin θ = 3/5, find cos θ. (ii) Find sin²θ + cos²θ. (iii) If tan θ = 4/3, find sec θ. (iv) Verify: 1 + tan²θ = sec²θ. Ans: (i) cos θ = 4/5. (ii) 9/25 + 16/25 = 1. (iii) sec θ = 5/3. (iv) 1 + 16/9 = 25/9 = sec²θ ✓ Amitesh Nagar, Indore (M.P.) PYQ SUMMARY & ANALYSIS Topic Years Asked Frequency Marks Standard angle values 2019–2025 Every Year 1–2 Trigonometric ratios from given info 2019–2024 Every Year 1–2 Identity: sin²θ + cos²θ = 1 2019–2025 Every Year 1–3 Prove that (identity proofs) 2019–2024 Every Year 3 Expression evaluation 2019–2024 5 times 1–2 Relation between ratios 2020–2024 4 times 2–3 sec²θ − tan²θ / cosec²θ − cot²θ 2019–2025 5 times 1–3 Case Study (triangle ratios) 2024–2025 2 times 4 Key Observations for Students:

"Prove that" identity questions are the MOST important — 3 marks, asked every year. Standard angle values (sin, cos, tan of 0°, 30°, 45°, 60°, 90°) — MUST MEMORIZE. Three identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ. Tip for proving: Convert everything to sin θ and cos θ, then simplify. Complementary angle ratios (sin(90°−θ) = cos θ etc.) are DELETED from 2025–26 syllabus. Expected marks from this chapter: 6–8 marks (combined with Ch 9). "Practice makes perfect. Solve PYQs to master your Board Exam!" Best Wishes for Your Board Exam!