Keerthana Classes

Maths ch8 introduction to trigonometry

October 04, 2026

37 marks. Read the notes, then tap Start test. Answers and your score appear only after you submit.

Quick notes

In a right triangle with acute angle A: sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse, tan A = opposite/adjacent.
cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A, and tan A = sin A/cos A.

Standard values: sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, sin 90° = 1.
cos takes the same values in reverse order. tan 0° = 0, tan 30° = 1/√3, tan 45° = 1, tan 60° = √3, tan 90° is not defined.

Identities: sin²A + cos²A = 1, 1 + tan²A = sec²A, 1 + cot²A = cosec²A.

Part A: Multiple choice 16 questions

1. sin 30° equals

  1. √3/2
  2. 1/2
  3. 1
  4. 1/√2

Standard value.

2. tan 45° equals

  1. 0
  2. √3
  3. 1
  4. 1/√3

In a 45° right triangle, opposite = adjacent.

3. cos 60° equals

  1. 1/2
  2. √3/2
  3. 1
  4. 0

cos 60° = sin 30° = 1/2.

4. sin²A + cos²A equals

  1. 0
  2. tan²A
  3. 2
  4. 1

This is the basic identity.

5. If tan A = 4/3 (A acute), then sin A is

  1. 3/5
  2. 4/5
  3. 3/4
  4. 5/4

Sides 4 and 3 give hypotenuse 5.

6. 1 + tan²A equals

  1. sec²A
  2. cosec²A
  3. cot²A
  4. sin²A

Divide sin²A + cos²A = 1 by cos²A.

7. (2 tan 30°)/(1 + tan²30°) equals

  1. sin 60°
  2. cos 60°
  3. tan 60°
  4. sin 30°

(2/√3)/(4/3) = √3/2 = sin 60°.

8. If sin A = 3/4, then cos A is

  1. 1/4
  2. 4/3
  3. √7/4
  4. 7/4

cos A = √(1 − 9/16) = √7/4.

9. sec A equals

  1. 1/sin A
  2. 1/tan A
  3. sin A/cos A
  4. 1/cos A

sec is the reciprocal of cos.

10. sin 60° cos 30° + sin 30° cos 60° equals

  1. 0
  2. 1
  3. 3/4
  4. 1/2

3/4 + 1/4 = 1.

11. 9 sec²A − 9 tan²A equals

  1. 1
  2. 0
  3. 9
  4. 8

9(sec²A − tan²A) = 9 × 1.

12. tan 90° is

  1. 0
  2. 1
  3. not defined
  4. √3

cos 90° = 0, so sin 90°/cos 90° is undefined.

13. If sec θ = 13/12, then cos θ is

  1. 12/13
  2. 5/13
  3. 13/12
  4. 5/12

cos θ = 1/sec θ.

14. In ΔABC right-angled at B, AB = 24 cm and BC = 7 cm. sin A is

  1. 24/25
  2. 7/24
  3. 7/25
  4. 25/7

AC = 25, and the side opposite A is BC = 7.

15. cos 0° equals

  1. 0
  2. 1/2
  3. √3/2
  4. 1

Standard value.

16. sin A = cos A when A is

  1. 30°
  2. 45°
  3. 60°
  4. 90°

Both equal 1/√2 at 45°.

Part B: Fill in the blanks 8 blanks

17. sin 90° = [[1]] and cos 90° = [[0]].
18. tan A = sin A ÷ [[cos A|cosa|cos]].
19. cosec A is the reciprocal of [[sin A|sina|sin|sine]].
20. sin²30° + cos²30° = [[1]].
21. If sin A = 1/2 and A is acute, then A = [[30]]°.
22. tan 45° + cot 45° = [[2]].
23. cos 0° × sin 90° = [[1]].

Part C: Questions and answers 9 questions

24. Given tan A = 4/3, find the other trigonometric ratios of A.

Opposite = 4k, adjacent = 3k, hypotenuse = 5k.
sin A = 4/5, cos A = 3/5, cosec A = 5/4, sec A = 5/3, cot A = 3/4.

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

AC = √(576 + 49) = 25.
sin A = 7/25, cos A = 24/25, sin C = 24/25, cos C = 7/25.

26. Evaluate 2 tan²45° + cos²30° − sin²60°.

2 × 1 + 3/4 − 3/4 = 2.

27. If sin(A − B) = 1/2 and cos(A + B) = 1/2, with 0° < A + B ≤ 90° and A > B, find A and B.

A − B = 30° and A + B = 60°.
Adding: 2A = 90°, so A = 45°, B = 15°.

28. Prove that (1 + tan²A)/(1 + cot²A) = tan²A.

LHS = sec²A/cosec²A = (1/cos²A) × sin²A = sin²A/cos²A = tan²A = RHS.

29. Prove that (sec A + tan A)(1 − sin A) = cos A.

LHS = [(1 + sin A)/cos A] × (1 − sin A) = (1 − sin²A)/cos A = cos²A/cos A = cos A = RHS.

30. If 15 cot A = 8, find sin A and sec A.

cot A = 8/15, so adjacent = 8, opposite = 15, hypotenuse = 17.
sin A = 15/17, sec A = 17/8.

31. Prove that (1 − cos²A) cosec²A = 1.

1 − cos²A = sin²A, so LHS = sin²A × 1/sin²A = 1 = RHS.

32. Evaluate (sin 30° + tan 45° − cosec 60°)/(sec 30° + cos 60° + cot 45°).

Numerator = 1/2 + 1 − 2/√3 = 3/2 − 2/√3. Denominator = 2/√3 + 1/2 + 1 = 3/2 + 2/√3.
Answer = (3√3 − 4)/(3√3 + 4).

Part D: Word puzzles 4 puzzles

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