Class 10 Mathematics: Introduction to Trigonometry important questions

Introduction to Trigonometry is a chapter of Class 10 Mathematics for the Andhra Pradesh State Board. These 17 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.

1. Practice · 1 marks

   1. Match: A) tan θ B) cot θ C) cosec θ with i) cos θ / sin θ ii) √(1 + cot²θ) iii) √(sec²θ − 1) a) A–i, B–ii, C–iii b) A–ii, B–iii, C–i c) A–iii, B–i, C–ii d) A–ii, B–i, C–iii

   c) tan θ = √(sec²θ − 1), cot θ = cos θ/sin θ, cosec θ = √(1 + cot²θ).

2. Important · 2 marks

   2. If tan θ = 7/24, find sec θ.

   sec²θ = 1 + tan²θ = 1 + 49/576 = 625/576, so sec θ = 25/24 (θ acute).

3. Practice · 1 marks

   3. If 4 cot A = 3, then tan A = ? A) 3/5 B) 4/5 C) 4/3 D) 3/4

   C) 4/3. cot A = 3/4, and tan A = 1/cot A = 4/3.

4. Important · 2 marks

   4. Express cos A and tan A in terms of sin A.

   From sin²A + cos²A = 1: cos A = √(1 − sin²A). tan A = sin A / cos A = sin A / √(1 − sin²A).

5. Important · 4 marks

   5. Prove that (cot A − cos A)/(cot A + cos A) = (cosec A − 1)/(cosec A + 1).

   LHS: write cot A = cos A / sin A and divide the numerator and denominator by cos A: (cot A − cos A)/(cot A + cos A) = (1/sin A − 1)/(1/sin A + 1) = (cosec A − 1)/(cosec A + 1) = RHS. Hence proved.

6. Practice · 1 marks

   6. Which trigonometric ratio equals (side adjacent to θ)/hypotenuse? A) sin θ B) cos θ C) tan θ D) cot θ

   B) cos θ

7. Important · 2 marks

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

   tan(A + B) = √3 = tan 60°, so A + B = 60°. tan(A − B) = 1/√3 = tan 30°, so A − B = 30°. Adding: 2A = 90°, so A = 45° (and B = 15°).

8. Important · 4 marks

   8. State the three trigonometric identities.

   1) sin²A + cos²A = 1 2) 1 + tan²A = sec²A (for 0° ≤ A < 90°) 3) 1 + cot²A = cosec²A (for 0° < A ≤ 90°)

9. Practice · 1 marks

   9. In a right triangle ABC, if tan A = 1, then ∠A = ? A) 0° B) 30° C) 45° D) 60°

   C) 45°, because tan 45° = 1.

10. Important · 2 marks

    10. If cot θ = 7/8, evaluate [(1 + sin θ)(1 − sin θ)] / [(1 + cos θ)(1 − cos θ)].

    (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.

11. Important · 4 marks

    11. Find the values of i) sin 60° cos 30° + sin 30° cos 60° ii) 2tan²45° + cos30° − sin²60°

    i) (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. ii) As printed: 2(1)² + √3/2 − (√3/2)² = 2 + √3/2 − 3/4 = (5 + 2√3)/4. (If the question meant cos²30°, as in the textbook: 2 + 3/4 − 3/4 = 2.)

12. Practice · 1 marks

    12. Ravi and his friend disagree while comparing the values of sin 27° and sin 72°. How would you help them resolve the conflict? A) Ask them to stop discussing the problem and move to another question. B) Explain that sine is an increasing function for angles between 0° and 90°, so sin 72° > sin 27°. C) Tell Ravi to accept his friend's answer without any explanation. D) Suggest that both answers are correct to avoid an argument.

    B) As the angle increases from 0° to 90°, sin increases from 0 to 1, so sin 72° > sin 27°.

13. Important · 2 marks

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

    sin(A + B) = 1 = sin 90°, so A + B = 90°. tan(A − B) = 1/√3 = tan 30°, so A − B = 30°. Adding: 2A = 120°, so A = 60° and B = 30°.

14. Important · 4 marks

    14. State whether the following statements are true or false. Justify your answer. A) sin(A + B) = sin A + sin B B) sin θ = cos θ for all values of θ

    A) False. Take A = B = 30°: sin 60° = √3/2 ≈ 0.87, but sin 30° + sin 30° = 1/2 + 1/2 = 1. B) False. It holds only for θ = 45° (in 0° to 90°). For θ = 0°, sin 0° = 0 but cos 0° = 1.

15. Practice · 1 marks

    15. Your classmate writes cos θ = (side adjacent to θ)/(side opposite to θ). Do you agree? If not, state the correct ratio.

    No. cos θ = (side adjacent to θ)/(hypotenuse). The ratio adjacent/opposite is cot θ.

16. Important · 2 marks

    16. If sec θ + tan θ = P, find sec θ in terms of P.

    sec²θ − tan²θ = 1, so (sec θ + tan θ)(sec θ − tan θ) = 1 and sec θ − tan θ = 1/P. Adding the two equations: 2 sec θ = P + 1/P. sec θ = (P² + 1)/(2P).

17. Important · 4 marks

    17. In ΔABC, right-angled at B, AB = 24 cm and BC = 7 cm. Find i) sin A, cos A ii) sin C, cos C

    AC = √(24² + 7²) = √(576 + 49) = √625 = 25 cm. i) sin A = BC/AC = 7/25, cos A = AB/AC = 24/25. ii) sin C = AB/AC = 24/25, cos C = BC/AC = 7/25.