Class 10 Mathematics: Introduction to Trigonometry important questions
Introduction to Trigonometry is a chapter of Class 10 Mathematics for the CBSE. These 24 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.
Practice · 1 marks
1. If x tan 60° cos 60° = sin 60° cot 60°, then x = (a) cos 30° (b) tan 30° (c) sin 30° (d) cot 30°
(b) tan 30° x × √3 × ½ = (√3/2)(1/√3) = ½, so x = 1/√3 = tan 30°.
Practice · 1 marks
2. If sin θ + cos θ = √2, then tan θ + cot θ = (a) 1 (b) 2 (c) 3 (d) 4
(b) 2 Squaring: 1 + 2 sin θ cos θ = 2, so sin θ cos θ = ½. tan θ + cot θ = 1/(sin θ cos θ) = 2.
Practice · 1 marks
3. If 5 tan β = 4, then (5 sin β − 2 cos β)/(5 sin β + 2 cos β) = (a) 1/3 (b) 2/5 (c) 3/5 (d) 6
(a) 1/3 Divide by cos β: (5 tan β − 2)/(5 tan β + 2) = (4 − 2)/(4 + 2) = 1/3.
Important · 2 marks
4. If sin(A + B) = 1 and cos(A − B) = √3/2, 0° < A + B ≤ 90°, A > B, find A and B.
A + B = 90° and A − B = 30°, so A = 60° and B = 30°.
Important · 2 marks
5. Find an acute angle θ when (cos θ − sin θ)/(cos θ + sin θ) = (1 − √3)/(1 + √3).
Divide by cos θ: (1 − tan θ)/(1 + tan θ) = (1 − √3)/(1 + √3). Comparing, tan θ = √3, so θ = 60°.
Important · 3 marks
6. Prove that tan³θ/(1 + tan²θ) + cot³θ/(1 + cot²θ) = sec θ cosec θ − 2 sin θ cos θ.
tan³θ/sec²θ = sin³θ/cos θ and cot³θ/cosec²θ = cos³θ/sin θ. LHS = (sin⁴θ + cos⁴θ)/(sin θ cos θ) = [(sin²θ + cos²θ)² − 2sin²θcos²θ]/(sin θ cos θ) = 1/(sin θ cos θ) − 2 sin θ cos θ = sec θ cosec θ − 2 sin θ cos θ = RHS.
Practice · 1 marks
7. Given that sin θ = a/b, cos θ is: (a) b/√(b² − a²) (b) b/a (c) √(b² − a²)/b (d) a/√(b² − a²)
(c) √(b² − a²)/b The opposite side is a and the hypotenuse b, so the adjacent side is √(b² − a²).
Practice · 1 marks
8. (sec A + tan A)(1 − sin A) equals: (a) sec A (b) sin A (c) cosec A (d) cos A
(d) cos A [(1 + sin A)/cos A](1 − sin A) = (1 − sin²A)/cos A = cos²A/cos A = cos A.
Important · 2 marks
9. If tan(A + B) = √3 and tan(A − B) = 1/√3, 0° < A + B ≤ 90°, A > B, find A and B.
A + B = 60° and A − B = 30°, so A = 45° and B = 15°.
Important · 2 marks
10. Find x if 2 cosec²30° + x sin²60° − (3/4) tan²30° = 10.
2(2)² + x(3/4) − (3/4)(1/3) = 10, so 8 + 3x/4 − 1/4 = 10. 3x/4 = 9/4, so x = 3.
Important · 3 marks
11. If 1 + sin²θ = 3 sin θ cos θ, prove that tan θ = 1 or 1/2.
Divide by cos²θ: sec²θ + tan²θ = 3 tan θ, so 1 + 2tan²θ = 3 tan θ. 2tan²θ − 3 tan θ + 1 = 0, i.e. (tan θ − 1)(2 tan θ − 1) = 0, so tan θ = 1 or 1/2.
Practice · 1 marks
12. If tan θ = 5/2, then (4 sin θ + cos θ)/(4 sin θ − cos θ) is equal to: (A) 11/9 (B) 3/2 (C) 9/11 (D) 4
(A) 11/9 Divide by cos θ: (4 tan θ + 1)/(4 tan θ − 1) = (10 + 1)/(10 − 1) = 11/9.
Practice · 1 marks
13. If θ = 30°, then the value of 3 tan θ is: (A) 1 (B) 1/√3 (C) 3/√3 (D) not defined
(C) 3/√3 (= √3), since tan 30° = 1/√3.
Important · 2 marks
14. Evaluate: (2 sin²60° − tan²30°)/sec²45°
= [2 × (√3/2)² − (1/√3)²] / (√2)² = (3/2 − 1/3)/2 = (7/6)/2 = 7/12.
Important · 3 marks
15. If cos θ + sin θ = 1, prove that cos θ − sin θ = ±1.
(cos θ + sin θ)² + (cos θ − sin θ)² = 2(cos²θ + sin²θ) = 2. So 1 + (cos θ − sin θ)² = 2, giving (cos θ − sin θ)² = 1 and cos θ − sin θ = ±1.
Practice · 1 marks
16. If sec θ + tan θ = x, then sec θ − tan θ is: (A) x (B) x² (C) 2/x (D) 1/x
(D) 1/x (sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ = 1, so sec θ − tan θ = 1/x.
Practice · 1 marks
17. If 2 sin 5x = √3, 0° ≤ x ≤ 90°, then x is: (A) 10° (B) 12° (C) 20° (D) 50°
(B) 12° sin 5x = √3/2, so 5x = 60° and x = 12°.
Practice · 1 marks
18. Assertion (A): If cos A + cos²A = 1, then sin²A + sin⁴A = 1. Reason (R): sin²A + cos²A = 1. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.
A. Both are true, and R explains A: cos A = 1 − cos²A = sin²A, so cos A + cos²A = 1 becomes sin²A + sin⁴A = 1.
Important · 2 marks
19. If sin(A + B) = 1 and cos(A − B) = √3/2, 0° < A, B < 90°, find A and B.
A + B = 90° and A − B = 30°, so A = 60° and B = 30°.
Important · 3 marks
20. If sin θ + cos θ = √3, prove that tan θ + cot θ = 1.
Squaring: 1 + 2 sin θ cos θ = 3, so sin θ cos θ = 1. tan θ + cot θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/1 = 1.
Important · 3 marks
21. Prove that (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A.
Divide the numerator and denominator by sin A: (cot A − 1 + cosec A)/(cot A + 1 − cosec A). Write 1 = cosec²A − cot²A in the numerator: (cot A + cosec A) − (cosec²A − cot²A) = (cosec A + cot A)(1 − cosec A + cot A). The denominator is (cot A − cosec A + 1), so they cancel, leaving cosec A + cot A.
Practice · 1 marks
22. If tan(A + B) = √3 and tan(A − B) = 1/√3, with 0° < A + B < 90° and A > B, then A and B are: (A) 60°, 30° (B) 60°, 45° (C) 45°, 15° (D) 60°, 15°
(C) 45°, 15° A + B = 60° and A − B = 30°, so A = 45° and B = 15°.
Important · 2 marks
23. If tan θ = 1/√5, find the value of (cosec²θ − sec²θ)/(cosec²θ + sec²θ).
cot θ = √5. sec²θ = 1 + tan²θ = 1 + 1/5 = 6/5, and cosec²θ = 1 + cot²θ = 1 + 5 = 6. Value = (6 − 6/5)/(6 + 6/5) = (24/5)/(36/5) = 2/3.
Important · 3 marks
24. Ms Isha asked the class to simplify cos θ/(1 − sin θ) + (1 − sin θ)/cos θ. Jyoti wrote: Step 1: [cos²θ + (1 − sin θ)²] / [(1 − sin θ)cos θ] Step 2: [cos²θ + cos²θ] / [(1 − sin θ)cos θ] Step 3: 2cos²θ / [(1 − sin θ)cos θ] Step 4: 2cos θ/(1 − sin θ) Find the step with the error, correct it, and find the right answer.
The error is in Step 2: (1 − sin θ)² is not cos²θ. Correct working: [cos²θ + 1 − 2sin θ + sin²θ] / [(1 − sin θ)cos θ] = (2 − 2sin θ) / [(1 − sin θ)cos θ] = 2(1 − sin θ) / [(1 − sin θ)cos θ] = 2/cos θ = 2 sec θ.
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