Important Questions Class 10 Mathematics Chapter 8 Introduction to Trigonometry

Trigonometry is the branch of mathematics that studies relationships between the sides and angles of a triangle. In a right triangle, sine, cosine, tangent, cosecant, secant, and cotangent connect an acute angle with side ratios.

Right triangles can measure heights, widths, and distances that students cannot measure directly. Important Questions Class 10 Mathematics Chapter 8 help students practise trigonometric ratios, reciprocal ratios, ratios from one given value, values of 0°, 30°, 45°, 60°, 90°, and trigonometric identities. The 2026 NCERT chapter Introduction to Trigonometry defines trigonometry as the study of relationships between sides and angles of a triangle, with discussion restricted mainly to acute angles.

Key Takeaways

• Trigonometric Ratios: sin A, cos A, and tan A compare opposite, adjacent, and hypotenuse sides.
• Reciprocal Ratios: cosec A, sec A, and cot A are reciprocals of sin A, cos A, and tan A.
• Specific Angles: The chapter uses 0°, 30°, 45°, 60°, and 90° values for direct evaluation questions.
• Trigonometric Identities: sin² A + cos² A = 1, sec² A − tan² A = 1, and cosec² A − cot² A = 1.

Important Questions Class 10 Mathematics Chapter 8 Structure 2026

Introduction to Trigonometry Class 10 Chapter Overview

Introduction to Trigonometry Class 10 begins with right-triangle situations like Qutub Minar, river width, and balloon altitude. These examples show why side-angle relationships matter in measurement.

The chapter then defines six trigonometric ratios, values of standard angles, and identities. NCERT also explains that trigonometric ratios of an angle do not change when triangle size changes, if the angle remains the same.

Class 10 Mathematics Chapter 8 Important Questions On Trigonometric Ratios

A right triangle has three side names based on the chosen angle. The opposite and adjacent sides change when the angle changes.

Students should label the angle first. Then they should identify opposite, adjacent, and hypotenuse.

Q1. What Are Trigonometric Ratios Class 10?

Trigonometric ratios are ratios of sides of a right triangle with respect to an acute angle.

They connect angle measure with side lengths. The six ratios are sin, cos, tan, cosec, sec, and cot.

For angle A:
sin A = opposite side/hypotenuse
cos A = adjacent side/hypotenuse
tan A = opposite side/adjacent side

Q2. In A Right Triangle ABC, Right-Angled At B, Write sin A, cos A And tan A.

For angle A, sin A = BC/AC, cos A = AB/AC, and tan A = BC/AB.

1. Hypotenuse = AC
2. Side opposite angle A = BC
3. Side adjacent to angle A = AB
4. sin A = BC/AC
5. cos A = AB/AC
6. tan A = BC/AB

Final Result: sin A = BC/AC, cos A = AB/AC, tan A = BC/AB

Q3. In The Same Triangle, Write sin C, cos C And tan C.

For angle C, sin C = AB/AC, cos C = BC/AC, and tan C = AB/BC.

1. Hypotenuse = AC
2. Side opposite angle C = AB
3. Side adjacent to angle C = BC
4. sin C = AB/AC
5. cos C = BC/AC
6. tan C = AB/BC

Final Result: sin C = AB/AC, cos C = BC/AC, tan C = AB/BC

Q4. Why Do Trigonometric Ratios Not Change When Triangle Size Changes?

Trigonometric ratios remain the same because similar right triangles have proportional corresponding sides.

If the angle stays fixed, the new right triangle stays similar to the original triangle. Corresponding side ratios remain equal.

Example: sin A = opposite/hypotenuse stays constant for the same angle A.

Q5. Why Are sin A And cos A Never Greater Than 1?

sin A and cos A never exceed 1 because the hypotenuse is the longest side.

sin A = opposite/hypotenuse and cos A = adjacent/hypotenuse. A smaller side divided by the hypotenuse cannot exceed 1.

Example: In a 3-4-5 triangle, sin A can be 3/5 or 4/5.

Sin Cos Tan Class 10 Questions With Answers

sin, cos, and tan form the base of most Chapter 8 questions. Once these three are known, the reciprocal ratios follow directly.

Use Pythagoras theorem whenever one side is missing.

Q6. In ∆ABC, Right-Angled At B, AB = 24 cm And BC = 7 cm. Find sin A And cos A.

sin A = 7/25 and cos A = 24/25.

1. Given Data:
   AB = 24 cm, BC = 7 cm, ∠B = 90°
2. Formula Used:
   AC² = AB² + BC²
3. Calculation:
   AC² = 24² + 7²
   AC² = 576 + 49 = 625
   AC = 25 cm
4. Ratios for angle A:
   sin A = BC/AC = 7/25
   cos A = AB/AC = 24/25

Final Result: sin A = 7/25, cos A = 24/25

Q7. In ∆ABC, Right-Angled At B, AB = 24 cm And BC = 7 cm. Find sin C And cos C.

sin C = 24/25 and cos C = 7/25.

1. Given Data:
   AB = 24 cm, BC = 7 cm, AC = 25 cm
2. For angle C:
   Opposite side = AB
   Adjacent side = BC
   Hypotenuse = AC
3. Calculation:
   sin C = AB/AC = 24/25
   cos C = BC/AC = 7/25

Final Result: sin C = 24/25, cos C = 7/25

Q8. In A Right Triangle, tan A = 4/3. Find sin A And cos A.

sin A = 4/5 and cos A = 3/5.

1. Given Data:
   tan A = opposite/adjacent = 4/3
2. Let:
   Opposite side = 4k
   Adjacent side = 3k
3. Use Pythagoras theorem:
   Hypotenuse² = (4k)² + (3k)²
   Hypotenuse² = 16k² + 9k² = 25k²
   Hypotenuse = 5k
4. Find ratios:
   sin A = 4k/5k = 4/5
   cos A = 3k/5k = 3/5

Final Result: sin A = 4/5, cos A = 3/5

Q9. If sin A = 3/4, Find cos A And tan A.

cos A = √7/4 and tan A = 3/√7.

1. Given Data:
   sin A = opposite/hypotenuse = 3/4
2. Let:
   Opposite side = 3k
   Hypotenuse = 4k
3. Use Pythagoras theorem:
   Adjacent² = (4k)² − (3k)²
   Adjacent² = 16k² − 9k² = 7k²
   Adjacent = √7 k
4. Find ratios:
   cos A = adjacent/hypotenuse = √7 k/4k = √7/4
   tan A = opposite/adjacent = 3k/√7 k = 3/√7

Final Result: cos A = √7/4, tan A = 3/√7

Q10. Given 15 cot A = 8, Find sin A And sec A.

sin A = 15/17 and sec A = 17/8.

1. Given Data:
   15 cot A = 8
   cot A = 8/15
2. Use cot A:
   cot A = adjacent/opposite = 8/15
3. Let:
   Adjacent side = 8k
   Opposite side = 15k
4. Hypotenuse:
   Hypotenuse² = (8k)² + (15k)²
   Hypotenuse² = 64k² + 225k² = 289k²
   Hypotenuse = 17k
5. Ratios:
   sin A = opposite/hypotenuse = 15/17
   sec A = hypotenuse/adjacent = 17/8

Final Result: sin A = 15/17, sec A = 17/8

Cosec Sec Cot Class 10 Important Questions

cosec, sec, and cot are reciprocal ratios. They become easy once sin, cos, and tan are clear.

NCERT also warns that sin A is not “sin × A”; it is the sine of angle A.

Q11. What Are cosec A, sec A And cot A?

cosec A, sec A, and cot A are reciprocal trigonometric ratios.

cosec A = 1/sin A
sec A = 1/cos A
cot A = 1/tan A

Example: If sin A = 3/5, then cosec A = 5/3.

Q12. Given sec θ = 13/12, Calculate All Other Trigonometric Ratios.

sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5.

1. Given Data:
   sec θ = 13/12
2. Convert to cos θ:
   cos θ = 1/sec θ = 12/13
3. Let:
   Adjacent side = 12k
   Hypotenuse = 13k
4. Find opposite side:
   Opposite² = (13k)² − (12k)²
   Opposite² = 169k² − 144k² = 25k²
   Opposite = 5k
5. Ratios:
   sin θ = 5/13
   tan θ = 5/12
   cosec θ = 13/5
   cot θ = 12/5

Final Result: sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5

Q13. If cot θ = 7/8, Evaluate cot² θ.

cot² θ = 49/64.

1. Given Data:
   cot θ = 7/8
2. Formula Used:
   cot² θ = (cot θ)²
3. Calculation:
   cot² θ = (7/8)²
   cot² θ = 49/64

Final Result: cot² θ = 49/64

Q14. If cot θ = 7/8, Evaluate (1 − sin² θ)/(1 − cos² θ).

The value is cot² θ = 49/64.

1. Given Data:
   cot θ = 7/8
2. Use identities:
   1 − sin² θ = cos² θ
   1 − cos² θ = sin² θ
3. Simplify:
   (1 − sin² θ)/(1 − cos² θ) = cos² θ/sin² θ
   = cot² θ
4. Substitute:
   cot² θ = (7/8)² = 49/64

Final Result: 49/64

Trigonometric Ratios Of Specific Angles Class 10

Standard angle values reduce long calculations. Chapter 8 derives 30°, 45°, and 60° values from special right triangles.

The chapter defines sin 0° = 0, cos 0° = 1, sin 90° = 1, and cos 90° = 0.

Q15. Write The Trigonometry Table Class 10 For 0°, 30°, 45°, 60° And 90°.

The standard trigonometry table gives fixed values for sin, cos, tan, cosec, sec, and cot.

Q16. Evaluate sin 60° cos 30° + sin 30° cos 60°.

The value is 1.

1. Given Expression:
   sin 60° cos 30° + sin 30° cos 60°
2. Values Used:
   sin 60° = √3/2
   cos 30° = √3/2
   sin 30° = 1/2
   cos 60° = 1/2
3. Calculation:
   (√3/2 × √3/2) + (1/2 × 1/2)
   = 3/4 + 1/4
   = 1

Final Result: 1

Q17. Evaluate 2 tan² 45° + cos² 30° − sin² 60°.

The value is 2.

1. Given Expression:
   2 tan² 45° + cos² 30° − sin² 60°
2. Values Used:
   tan 45° = 1
   cos 30° = √3/2
   sin 60° = √3/2
3. Calculation:
   2(1)² + (√3/2)² − (√3/2)²
   = 2 + 3/4 − 3/4
   = 2

Final Result: 2

Q18. Evaluate cos 45°/(sec 30° + cosec 30°).

The value is 3√2/(4 + 2√3).

1. Given Expression:
   cos 45°/(sec 30° + cosec 30°)
2. Values Used:
   cos 45° = 1/√2
   sec 30° = 2/√3
   cosec 30° = 2
3. Substitute:
   (1/√2)/(2/√3 + 2)
4. Simplify:
   = 1/[√2(2/√3 + 2)]
   = 1/[(2√2/√3) + 2√2]

Final Result: 3√2/(4 + 2√3)

Q19. Find A And B If sin(A − B) = 1/2, cos(A + B) = 1/2, A > B.

A = 45° and B = 15°.

1. Given Data:
   sin(A − B) = 1/2
   cos(A + B) = 1/2
2. Standard values:
   sin 30° = 1/2
   cos 60° = 1/2
3. Form equations:
   A − B = 30°
   A + B = 60°
4. Add equations:
   2A = 90°
   A = 45°
5. Find B:
   A + B = 60°
   45° + B = 60°
   B = 15°

Final Result: A = 45°, B = 15°

Q20. If tan(A + B) = √3 And tan(A − B) = 1/√3, Find A And B.

A = 45° and B = 15°.

1. Given Data:
   tan(A + B) = √3
   tan(A − B) = 1/√3
2. Standard values:
   tan 60° = √3
   tan 30° = 1/√3
3. Form equations:
   A + B = 60°
   A − B = 30°
4. Add equations:
   2A = 90°
   A = 45°
5. Find B:
   A + B = 60°
   45° + B = 60°
   B = 15°

Final Result: A = 45°, B = 15°

Values Of Trigonometric Ratios Class 10 From Right Triangles

Some questions give one side and one angle. Students should pick the ratio that uses known and required sides.

The NCERT examples use 30° triangles and Pythagoras theorem for side lengths.

Q21. In ∆ABC, Right-Angled At B, AB = 5 cm And ∠ACB = 30°. Find BC And AC.

BC = 5√3 cm and AC = 10 cm.

1. Given Data:
   ∠B = 90°
   AB = 5 cm
   ∠C = 30°
2. To find BC:
   tan 30° = AB/BC
3. Substitute:
   1/√3 = 5/BC
4. Solve:
   BC = 5√3 cm
5. To find AC:
   sin 30° = AB/AC
6. Substitute:
   1/2 = 5/AC
   AC = 10 cm

Final Result: BC = 5√3 cm, AC = 10 cm

Q22. In ∆PQR, Right-Angled At Q, PQ = 3 cm And PR = 6 cm. Find ∠QPR And ∠PRQ.

∠PRQ = 30° and ∠QPR = 60°.

1. Given Data:
   ∠Q = 90°
   PQ = 3 cm
   PR = 6 cm
2. Use angle R:
   sin R = opposite/hypotenuse
   sin R = PQ/PR = 3/6 = 1/2
3. Therefore:
   R = 30°
4. Find angle P:
   P + R = 90°
   P = 90° − 30° = 60°

Final Result: ∠PRQ = 30°, ∠QPR = 60°

Q23. In ∆PQR, Right-Angled At Q, PR + QR = 25 cm And PQ = 5 cm. Find sin P, cos P And tan P.

sin P = 24/25, cos P = 7/25, and tan P = 24/7.

1. Given Data:
   PR + QR = 25 cm
   PQ = 5 cm
   Let QR = x cm
2. Then:
   PR = 25 − x
3. Use Pythagoras theorem:
   PR² = PQ² + QR²
   (25 − x)² = 5² + x²
4. Expand and solve:
   625 − 50x + x² = 25 + x²
   600 = 50x
   x = 12
5. So:
   QR = 12 cm
   PR = 13 cm
6. For angle P:
   Opposite = QR = 12 cm
   Adjacent = PQ = 5 cm
   Hypotenuse = PR = 13 cm
7. Ratios:
   sin P = 12/13
   cos P = 5/13
   tan P = 12/5

Final Result: sin P = 12/13, cos P = 5/13, tan P = 12/5

Trigonometric Identities Class 10 Important Questions

Trigonometric identities are equations true for all defined values of the angle. The main identities come from Pythagoras theorem.

NCERT proves sin² A + cos² A = 1 by dividing AB² + BC² = AC² by AC².

Q24. Write The Three Main Trigonometric Identities Class 10.

The three main identities are sin² A + cos² A = 1, 1 + tan² A = sec² A, and 1 + cot² A = cosec² A.

They come from Pythagoras theorem in a right triangle. Each identity has a defined angle range.

Identities:
sin² A + cos² A = 1
1 + tan² A = sec² A
1 + cot² A = cosec² A

Q25. Prove sin² A + cos² A = 1.

sin² A + cos² A = 1 follows from Pythagoras theorem.

1. In ∆ABC, right-angled at B:
   AB² + BC² = AC²
2. Divide each term by AC²:
   AB²/AC² + BC²/AC² = AC²/AC²
3. Rewrite ratios:
   (AB/AC)² + (BC/AC)² = 1
4. Use definitions:
   cos² A + sin² A = 1

Final Result: sin² A + cos² A = 1

Q26. Prove 1 + tan² A = sec² A.

1 + tan² A = sec² A follows by dividing Pythagoras theorem by AB².

1. In ∆ABC, right-angled at B:
   AB² + BC² = AC²
2. Divide each term by AB²:
   AB²/AB² + BC²/AB² = AC²/AB²
3. Rewrite ratios:
   1 + (BC/AB)² = (AC/AB)²
4. Use definitions:
   1 + tan² A = sec² A

Final Result: 1 + tan² A = sec² A

Q27. Prove 1 + cot² A = cosec² A.

1 + cot² A = cosec² A follows by dividing Pythagoras theorem by BC².

1. In ∆ABC, right-angled at B:
   AB² + BC² = AC²
2. Divide each term by BC²:
   AB²/BC² + BC²/BC² = AC²/BC²
3. Rewrite ratios:
   (AB/BC)² + 1 = (AC/BC)²
4. Use definitions:
   cot² A + 1 = cosec² A

Final Result: 1 + cot² A = cosec² A

Q28. If tan A = 1/√3, Find sin A, cos A And sec A Using Identities.

sin A = 1/2, cos A = √3/2, and sec A = 2/√3.

1. Given Data:
   tan A = 1/√3
2. Use identity:
   1 + tan² A = sec² A
3. Substitute:
   sec² A = 1 + (1/√3)²
   sec² A = 1 + 1/3 = 4/3
4. Therefore:
   sec A = 2/√3
   cos A = 1/sec A = √3/2
5. Use identity:
   sin² A + cos² A = 1
   sin² A = 1 − 3/4 = 1/4
   sin A = 1/2

Final Result: sin A = 1/2, cos A = √3/2, sec A = 2/√3

Prove Trigonometric Identities Class 10 Questions

Identity questions reward clean conversion. The best method is to convert sec, cosec, tan, and cot into sin and cos where needed.

Students should simplify one side and reach the other side exactly.

Q29. Prove sec A(1 − sin A)(sec A + tan A) = 1.

The identity is proved by writing sec A and tan A in terms of sin A and cos A.

1. Start with LHS:
   sec A(1 − sin A)(sec A + tan A)
2. Convert ratios:
   sec A = 1/cos A
   tan A = sin A/cos A
3. Substitute:
   LHS = (1/cos A)(1 − sin A)(1/cos A + sin A/cos A)
4. Combine bracket:
   1/cos A + sin A/cos A = (1 + sin A)/cos A
5. Multiply:
   LHS = [(1 − sin A)(1 + sin A)]/cos² A
6. Use identity:
   (1 − sin A)(1 + sin A) = 1 − sin² A = cos² A
7. Simplify:
   LHS = cos² A/cos² A = 1

Final Result: sec A(1 − sin A)(sec A + tan A) = 1

Q30. Prove (sec A + tan A)(1 − sin A) = cos A.

The identity is proved by converting sec A and tan A into sin A and cos A.

1. Start with LHS:
   (sec A + tan A)(1 − sin A)
2. Convert ratios:
   sec A = 1/cos A
   tan A = sin A/cos A
3. Substitute:
   LHS = (1/cos A + sin A/cos A)(1 − sin A)
4. Combine bracket:
   LHS = [(1 + sin A)/cos A](1 − sin A)
5. Multiply:
   LHS = (1 − sin² A)/cos A
6. Use identity:
   1 − sin² A = cos² A
7. Simplify:
   LHS = cos² A/cos A = cos A

Final Result: (sec A + tan A)(1 − sin A) = cos A

Q31. Prove (cosec A − cot A)² = (1 − cos A)/(1 + cos A).

The identity is proved by writing cosec A and cot A in sine and cosine form.

1. Start with LHS:
   (cosec A − cot A)²
2. Convert ratios:
   cosec A = 1/sin A
   cot A = cos A/sin A
3. Substitute:
   LHS = [(1 − cos A)/sin A]²
4. Square:
   LHS = (1 − cos A)²/sin² A
5. Use identity:
   sin² A = 1 − cos² A
   sin² A = (1 − cos A)(1 + cos A)
6. Substitute:
   LHS = (1 − cos A)²/[(1 − cos A)(1 + cos A)]
7. Cancel common factor:
   LHS = (1 − cos A)/(1 + cos A)

Final Result: (cosec A − cot A)² = (1 − cos A)/(1 + cos A)

Q32. Prove (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

The identity is proved by expanding both squares and using reciprocal products.

1. Start with LHS:
   (sin A + cosec A)² + (cos A + sec A)²
2. Expand:
   sin² A + cosec² A + 2sin A cosec A
   • cos² A + sec² A + 2cos A sec A
3. Use reciprocal products:
   sin A cosec A = 1
   cos A sec A = 1
4. Substitute:
   LHS = sin² A + cos² A + cosec² A + sec² A + 4
5. Use identities:
   sin² A + cos² A = 1
   cosec² A = 1 + cot² A
   sec² A = 1 + tan² A
6. Combine:
   LHS = 1 + (1 + cot² A) + (1 + tan² A) + 4
7. Simplify:
   LHS = 7 + tan² A + cot² A

Final Result: RHS proved

NCERT Class 10 Mathematics Chapter 8 Questions

NCERT questions often ask students to verify true-false statements. These test definitions, limits, and notation.

The chapter clearly states that cosec A, sec A, and cot A are reciprocal ratios, not new side names.

Q33. Is tan A Always Less Than 1?

No, tan A is not always less than 1.

tan A increases as angle A increases from 0° to 90°. At 45°, tan A = 1, and at 60°, tan A = √3.

Final Result: False

Q34. Can sec A = 12/5 For Some Angle A?

Yes, sec A = 12/5 is possible for an acute angle A.

sec A = hypotenuse/adjacent side. Since hypotenuse is greater than adjacent side, sec A can exceed 1.

Example: sec A = 12/5 is greater than 1.

Final Result: True

Q35. Is cos A The Abbreviation For Cosecant Of Angle A?

No, cos A is the abbreviation for cosine of angle A.

cosecant of angle A is written as cosec A. The two ratios have different meanings.

Final Result: False

Q36. Is cot A The Product Of cot And A?

No, cot A means cotangent of angle A.

The symbol cot has no separate numerical meaning without an angle. NCERT states similar interpretation for sin A and cos A.

Final Result: False

Q37. Can sin θ = 4/3 For Some Angle θ?

No, sin θ cannot be 4/3 for any acute angle θ.

sin θ = opposite/hypotenuse. The opposite side cannot be longer than the hypotenuse.

Final Result: False

Class 10 Maths Trigonometry Questions With Answers For Board Pattern

Board-style questions often combine a given ratio with identities. Students should draw a right triangle or convert ratios into sin and cos.

Use exact values like √2, √3, and fractions in final answers.

Q38. If 3 cot A = 4, Check Whether (1 − tan² A)/(1 + tan² A) = cos² A − sin² A.

Yes, both sides are equal.

1. Given Data:
   3 cot A = 4
   cot A = 4/3
   tan A = 3/4
2. Find LHS:
   (1 − tan² A)/(1 + tan² A)
   = (1 − 9/16)/(1 + 9/16)
   = (7/16)/(25/16)
   = 7/25
3. Find sin A and cos A:
   If tan A = 3/4, then opposite = 3k, adjacent = 4k
   Hypotenuse = 5k
   sin A = 3/5, cos A = 4/5
4. Find RHS:
   cos² A − sin² A
   = (4/5)² − (3/5)²
   = 16/25 − 9/25
   = 7/25

Final Result: LHS = RHS = 7/25

Q39. In ∆ABC, Right-Angled At B, If tan A = 1/√3, Find sin A cos C + cos A sin C.

The value is 1.

1. Given Data:
   ∠B = 90°
   tan A = 1/√3
2. Since tan 30° = 1/√3:
   A = 30°
3. In right triangle ABC:
   A + C = 90°
   C = 60°
4. Evaluate expression:
   sin A cos C + cos A sin C
   = sin 30° cos 60° + cos 30° sin 60°
5. Substitute values:
   = (1/2 × 1/2) + (√3/2 × √3/2)
   = 1/4 + 3/4
   = 1

Final Result: 1

Q40. In ∆ABC, Right-Angled At B, If tan A = 1/√3, Find cos A cos C − sin A sin C.

The value is 0.

1. Given Data:
   tan A = 1/√3
2. Therefore:
   A = 30°
   C = 60°
3. Evaluate expression:
   cos A cos C − sin A sin C
   = cos 30° cos 60° − sin 30° sin 60°
4. Substitute values:
   = (√3/2 × 1/2) − (1/2 × √3/2)
   = √3/4 − √3/4
   = 0

Final Result: 0

Important Questions Class 10 Maths All Chapters