Introduction to Trigonometry — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Introduction to Trigonometry" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Introduction to Trigonometry" — 9 important questions with detailed answers for CB…
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NCERT Solutions for Class 10 Mathematics — Introduction to Trigonometry. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Introduction to Trigonometry — All 9 Questions Solved
Q1. Define the six trigonometric ratios for a right triangle. State the trigonometric values for the standard angles 0°, 30°, 45°, 60°, and 90°.
Trigonometric Ratios (for a right triangle with reference angle θ):
Let a right triangle have:
- Opposite side (opposite to angle θ) = opposite
- Adjacent side (adjacent to angle θ) = adjacent
- Hypotenuse = hypotenuse
The six ratios are:
1. sin θ = opposite / hypotenuse
2. cos θ = adjacent / hypotenuse
3. tan θ = opposite / adjacent
4. cot θ = adjacent / opposite = 1/tan θ
5. sec θ = hypotenuse / adjacent = 1/cos θ
6. cosec θ = hypotenuse / opposite = 1/sin θ
MNEMONIC: SOH-CAH-TOA
Sin=Opposite/Hypotenuse, Cos=Adjacent/Hypotenuse, Tan=Opposite/Adjacent
═══════════════════════════════════════════════════════════════
TRIGONOMETRIC VALUES FOR STANDARD ANGLES
═══════════════════════════════════════════════════════════════
Angle θ | 0° | 30° | 45° | 60° | 90°
-----------|-------|-------|-------|-------|--------
sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1
cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0
tan θ | 0 | 1/√3 | 1 | √3 | ∞(undef)
cot θ | ∞ | √3 | 1 | 1/√3 | 0
sec θ | 1 | 2/√3 | √2 | 2 | ∞(undef)
cosec θ | ∞ | 2 | √2 | 2/√3 | 1
Note: 1/√2 = √2/2, and 1/√3 = √3/3 (rationalized forms)
Key Observations:
- As angle increases from 0° to 90°, sin increases from 0 to 1
- As angle increases from 0° to 90°, cos decreases from 1 to 0
- sin²θ + cos²θ = 1 (Fundamental Identity)
Q2. In a right triangle ABC with right angle at B, AB = 5 cm and AC = 13 cm. Find all six trigonometric ratios of angle A.
Given:
Right triangle ABC with ∠B = 90°
AB = 5 cm (side adjacent to angle A)
AC = 13 cm (hypotenuse)
Find: All six trigonometric ratios of angle A
Step 1: Find BC (the side opposite to angle A) using Pythagorean theorem.
AC² = AB² + BC²
13² = 5² + BC²
169 = 25 + BC²
BC² = 144
BC = 12 cm
Step 2: Identify the sides relative to angle A.
Opposite side (opposite to ∠A) = BC = 12 cm
Adjacent side (adjacent to ∠A) = AB = 5 cm
Hypotenuse = AC = 13 cm
Step 3: Calculate sin A.
sin A = opposite / hypotenuse = 12/13
Step 4: Calculate cos A.
cos A = adjacent / hypotenuse = 5/13
Step 5: Calculate tan A.
tan A = opposite / adjacent = 12/5
Step 6: Calculate cot A.
cot A = adjacent / opposite = 5/12
Step 7: Calculate sec A.
sec A = hypotenuse / adjacent = 13/5
Step 8: Calculate cosec A.
cosec A = hypotenuse / opposite = 13/12
Answer:
sin A = 12/13
cos A = 5/13
tan A = 12/5
cot A = 5/12
sec A = 13/5
cosec A = 13/12
Q3. Prove the identity: sin²θ + cos²θ = 1
Prove: sin²θ + cos²θ = 1
Proof:
Consider a right triangle with:
- Angle θ at one vertex
- Opposite side = a
- Adjacent side = b
- Hypotenuse = c
Step 1: Write the definitions of sin θ and cos θ.
sin θ = a/c
cos θ = b/c
Step 2: Square both ratios.
sin²θ = a²/c²
cos²θ = b²/c²
Step 3: Add the squared ratios.
sin²θ + cos²θ = a²/c² + b²/c²
sin²θ + cos²θ = (a² + b²)/c²
Step 4: Apply the Pythagorean theorem.
In a right triangle: a² + b² = c²
Substitute:
sin²θ + cos²θ = c²/c²
sin²θ + cos²θ = 1
Hence proved. ✓
This is called the Pythagorean identity and is one of the most important trigonometric identities.
Other useful identities derived from this:
1. 1 + tan²θ = sec²θ
2. 1 + cot²θ = cosec²θ
Q4. Simplify: (sin 30° + cos 60°)/(tan 45°) + sec 60° - cosec 30°
Given: Simplify (sin 30° + cos 60°)/(tan 45°) + sec 60° - cosec 30°
Step 1: Recall the values of trigonometric functions for standard angles.
sin 30° = 1/2
cos 60° = 1/2
tan 45° = 1
sec 60° = 2
cosec 30° = 2
Step 2: Substitute these values.
= (1/2 + 1/2)/(1) + 2 - 2
Step 3: Simplify the first term.
= (1)/(1) + 2 - 2
= 1 + 2 - 2
= 1
Answer: The simplified expression equals 1
Q5. If sin θ = 3/5 and θ is acute, find cos θ and tan θ.
Given: sin θ = 3/5, θ is acute (0° < θ < 90°)
Find: cos θ and tan θ
Step 1: Use the Pythagorean identity sin²θ + cos²θ = 1.
sin²θ + cos²θ = 1
(3/5)² + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 1 - 9/25
cos²θ = 25/25 - 9/25
cos²θ = 16/25
Step 2: Take the square root.
cos θ = ±4/5
Since θ is acute, cos θ > 0.
cos θ = 4/5
Step 3: Find tan θ.
tan θ = sin θ / cos θ
tan θ = (3/5) / (4/5)
tan θ = (3/5) × (5/4)
tan θ = 3/4
Answer:
cos θ = 4/5 = 0.8
tan θ = 3/4 = 0.75
Verification:
sin²θ + cos²θ = (3/5)² + (4/5)² = 9/25 + 16/25 = 25/25 = 1 ✓
Q6. If tan θ = 5/12 and θ is in the first quadrant, find sin θ and cos θ.
Given: tan θ = 5/12, θ is in the first quadrant
Find: sin θ and cos θ
Step 1: Understand what tan θ = 5/12 means.
tan θ = opposite / adjacent = 5/12
This means in a right triangle:
Opposite side = 5
Adjacent side = 12
Step 2: Find the hypotenuse using Pythagorean theorem.
hypotenuse² = opposite² + adjacent²
hypotenuse² = 5² + 12²
hypotenuse² = 25 + 144
hypotenuse² = 169
hypotenuse = 13
Step 3: Find sin θ.
sin θ = opposite / hypotenuse = 5/13
Step 4: Find cos θ.
cos θ = adjacent / hypotenuse = 12/13
Answer:
sin θ = 5/13
cos θ = 12/13
Verification:
sin²θ + cos²θ = (5/13)² + (12/13)² = 25/169 + 144/169 = 169/169 = 1 ✓
tan θ = sin θ / cos θ = (5/13) / (12/13) = 5/12 ✓
Q7. Prove that: (1 - sin²θ) × sec²θ = 1
Prove: (1 - sin²θ) × sec²θ = 1
Proof:
Method 1: Using Pythagorean Identity
Step 1: Simplify the first bracket using sin²θ + cos²θ = 1.
1 - sin²θ = cos²θ
Step 2: Substitute this into the expression.
(1 - sin²θ) × sec²θ = cos²θ × sec²θ
Step 3: Recall that sec θ = 1/cos θ, so sec²θ = 1/cos²θ.
= cos²θ × (1/cos²θ)
= cos²θ / cos²θ
= 1
Hence proved. ✓
Method 2: Direct substitution
Step 1: Start with the left side.
(1 - sin²θ) × sec²θ
Step 2: Use sec²θ = 1/cos²θ.
= (1 - sin²θ) × (1/cos²θ)
= (1 - sin²θ) / cos²θ
Step 3: Use sin²θ + cos²θ = 1, so 1 - sin²θ = cos²θ.
= cos²θ / cos²θ
= 1
Hence proved. ✓
Q8. Evaluate: 3sin 30° + 4cos 60° - 2tan 45° + sec 45°
Given: Evaluate 3sin 30° + 4cos 60° - 2tan 45° + sec 45°
Step 1: Recall the standard angle values.
sin 30° = 1/2
cos 60° = 1/2
tan 45° = 1
sec 45° = 1/cos 45° = 1/(1/√2) = √2
Step 2: Substitute these values into the expression.
= 3(1/2) + 4(1/2) - 2(1) + √2
Step 3: Calculate each term.
= 3/2 + 4/2 - 2 + √2
= 3/2 + 2 - 2 + √2
= 3/2 + √2
Step 4: Convert to decimal if needed.
= 1.5 + 1.414...
= 2.914... (approximately)
Answer: 3/2 + √2 or (3 + 2√2)/2
Q9. In a right triangle, one angle is 30°. If the side opposite to the 30° angle is 5 cm, find the hypotenuse and the other leg.
Given:
Right triangle with one angle = 30°
Side opposite to 30° = 5 cm
Find: Hypotenuse and the other leg
Setup:
In a right triangle, we have:
- Right angle (90°)
- Angle of 30°
- Third angle = 180° - 90° - 30° = 60°
Step 1: Find the hypotenuse using sin 30°.
sin 30° = opposite / hypotenuse
1/2 = 5 / hypotenuse
hypotenuse = 5 × 2
hypotenuse = 10 cm
Step 2: Find the other leg (adjacent to 30°) using tan 30°.
tan 30° = opposite / adjacent
1/√3 = 5 / adjacent
adjacent = 5 × √3
adjacent = 5√3 cm
Alternative method for Step 2:
Using Pythagorean theorem:
hypotenuse² = opposite² + adjacent²
10² = 5² + adjacent²
100 = 25 + adjacent²
adjacent² = 75
adjacent = √75 = √(25 × 3) = 5√3 cm
Answer:
Hypotenuse = 10 cm
Other leg (adjacent to 30°) = 5√3 cm ≈ 8.66 cm
Verification:
sin 30° = 5/10 = 1/2 ✓
tan 30° = 5/(5√3) = 1/√3 ✓
5² + (5√3)² = 25 + 75 = 100 = 10² ✓
Introduction to Trigonometry — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. In a right-angled triangle ABC, right-angled at B, if tan A = 1, then the value of sin A cos A is:
- A. 1
- B. 1/2
- C. 2
- D. √2
Answer: B. 1/2 — If tan A = 1, then angle A must be 45°. Therefore, sin A cos A = sin 45° cos 45° = (1/√2)(1/√2) = 1/2.
MCQ 2. The value of (sin 30° + cos 60°) is:
- A. 1/2
- B. 1
- C. √3/2
- D. 2
Answer: B. 1 — We know that sin 30° = 1/2 and cos 60° = 1/2. Adding these values, 1/2 + 1/2 = 1.
MCQ 3. If sin θ = 3/5, then the value of (tan θ + sec θ) is:
- A. 3/2
- B. 2
- C. 4/3
- D. 5/3
Answer: B. 2 — Given sin θ = 3/5. Using a right triangle (Perpendicular=3, Hypotenuse=5), Base = √(5²-3²) = 4. So, tan θ = 3/4 and sec θ = 5/4. Thus, tan θ + sec θ = 3/4 + 5/4 = 8/4 = 2.
MCQ 4. The value of sin 60° cos 30° + sin 30° cos 60° is:
- A. 1/2
- B. √3/2
- C. 1
- D. 0
Answer: C. 1 — Using the known values: (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. This is also the expansion of sin(60°+30°) = sin 90° = 1.
MCQ 5. If √3 tan θ = 1, then the value of (cos²θ - sin²θ) is:
- A. 1/2
- B. 1/3
- C. 1
- D. 0
Answer: A. 1/2 — Given √3 tan θ = 1, so tan θ = 1/√3. This implies θ = 30°. Then, cos²30° - sin²30° = (√3/2)² - (1/2)² = 3/4 - 1/4 = 2/4 = 1/2.
MCQ 6. What is the value of 9 sec²A - 9 tan²A?
- A. 1
- B. 9
- C. 8
- D. 0
Answer: B. 9 — Factor out 9: 9(sec²A - tan²A). Using the trigonometric identity sec²A - tan²A = 1, the expression simplifies to 9(1) = 9.
MCQ 7. If tan A = 3/4, then sin A equals:
- A. 4/5
- B. 3/5
- C. 5/3
- D. 5/4
Answer: B. 3/5 — If tan A = 3/4 (opposite/adjacent), then by Pythagoras theorem, the hypotenuse is √(3²+4²) = 5. So, sin A = opposite/hypotenuse = 3/5.
MCQ 8. The value of (1 - tan²45°) / (1 + tan²45°) is:
- A. 0
- B. 1
- C. 2
- D. 1/2
Answer: A. 0 — Since tan 45° = 1, substitute this value into the expression: (1 - 1²) / (1 + 1²) = (1 - 1) / (1 + 1) = 0/2 = 0.
MCQ 9. If sin (A+B) = 1 and cos (A-B) = √3/2, where 0° < A+B ≤ 90° and A > B, then the values of A and B are respectively:
- A. A=60°, B=30°
- B. A=45°, B=45°
- C. A=75°, B=15°
- D. A=90°, B=0°
Answer: A. A=60°, B=30° — From sin (A+B) = 1, we get A+B = 90°. From cos (A-B) = √3/2, we get A-B = 30°. Solving these two linear equations: (A+B)+(A-B)=120° => 2A=120° => A=60°. Substituting A=60° into A+B=90°, we find B=30°.
MCQ 10. Evaluate sin²60° + cos²30° - sin²45°.
- A. 1/2
- B. 1
- C. 3/4
- D. 5/4
Answer: B. 1 — sin²60° = (√3/2)² = 3/4. cos²30° = (√3/2)² = 3/4. sin²45° = (1/√2)² = 1/2. So, 3/4 + 3/4 - 1/2 = 6/4 - 1/2 = 3/2 - 1/2 = 1.
MCQ 11. The value of tan 0° is:
- A. 0
- B. 1
- C. undefined
- D. √3
Answer: A. 0 — The tangent of an angle is defined as sin(angle)/cos(angle). Since sin 0° = 0 and cos 0° = 1, tan 0° = 0/1 = 0.
MCQ 12. If cos θ = 1/2, then the value of (2 sec θ + 3 tan θ) is:
- A. 4+3√3
- B. 2+3√3
- C. 1+3√3
- D. 4+√3
Answer: A. 4+3√3 — If cos θ = 1/2, then θ = 60°. So, sec θ = 1/cos θ = 2, and tan θ = tan 60° = √3. Therefore, 2 sec θ + 3 tan θ = 2(2) + 3(√3) = 4 + 3√3.
MCQ 13. If tan A = n tan B and sin A = m sin B, then cos²A is:
- A. (m²-1)/(n²-1)
- B. (m²+1)/(n²+1)
- C. (n²-1)/(m²-1)
- D. (n²+1)/(m²+1)
Answer: A. (m²-1)/(n²-1) — From sin A = m sin B, sin B = sin A / m. From tan A = n tan B, (sin A/cos A) = n (sin B/cos B). Substitute sin B: (sin A/cos A) = n (sin A / (m cos B)) => cos B = n cos A / m. Using sin²B + cos²B = 1: (sin A / m)² + (n cos A / m)² = 1. This simplifies to sin²A + n²cos²A = m². Replacing sin²A with (1 - cos²A), we get (1 - cos²A) + n²cos²A = m², so (n² - 1)cos²A = m² - 1. Thus, cos²A = (m² - 1) / (n² - 1).
MCQ 14. Evaluate sin 25° cos 65° + cos 25° sin 65°.
- A. 0
- B. 1
- C. 1/2
- D. √3/2
Answer: B. 1 — This expression is in the form sin A cos B + cos A sin B, which is equal to sin(A+B). So, sin(25° + 65°) = sin 90° = 1.
MCQ 15. If tan (A+B) = √3 and tan (A-B) = 1/√3, where 0° < A+B ≤ 90° and A > B, then the values of A and B are respectively:
- A. A=45°, B=15°
- B. A=60°, B=30°
- C. A=30°, B=15°
- D. A=45°, B=30°
Answer: A. A=45°, B=15° — From tan (A+B) = √3, we get A+B = 60°. From tan (A-B) = 1/√3, we get A-B = 30°. Adding these two equations gives 2A = 90°, so A = 45°. Substituting A=45° into A+B=60° yields B = 15°.
MCQ 16. The value of (sin 70° / cos 20°) is:
- A. 0
- B. 1
- C. -1
- D. 2
Answer: B. 1 — Since 70° and 20° are complementary angles (70° + 20° = 90°), we know that cos 20° = sin (90° - 20°) = sin 70°. Therefore, sin 70° / cos 20° = sin 70° / sin 70° = 1.
MCQ 17. If x = a cos θ and y = b sin θ, then what is the relationship between x, y, a, b?
- A. x²/a² + y²/b² = 1
- B. x²/b² + y²/a² = 1
- C. a²/x² + b²/y² = 1
- D. x²a² + y²b² = 1
Answer: A. x²/a² + y²/b² = 1 — From x = a cos θ, we get cos θ = x/a. From y = b sin θ, we get sin θ = y/b. Using the identity sin²θ + cos²θ = 1, we substitute the values: (y/b)² + (x/a)² = 1, which simplifies to x²/a² + y²/b² = 1.
MCQ 18. If 2 sin 2θ = √3, then the value of θ is:
- A. 30°
- B. 45°
- C. 60°
- D. 15°
Answer: A. 30° — Given 2 sin 2θ = √3, so sin 2θ = √3/2. We know that sin 60° = √3/2. Therefore, 2θ = 60°, which implies θ = 30°.
MCQ 19. If 5 cot θ = 12, then sin θ is:
- A. 5/12
- B. 12/13
- C. 5/13
- D. 12/5
Answer: C. 5/13 — If 5 cot θ = 12, then cot θ = 12/5 (adjacent/opposite). In a right-angled triangle, if adjacent=12 and opposite=5, the hypotenuse = √(12²+5²) = √169 = 13. Therefore, sin θ = opposite/hypotenuse = 5/13.
MCQ 20. Simplify (sin θ - 2 sin³θ) / (2 cos³θ - cos θ).
- A. tan θ
- B. cot θ
- C. sec θ
- D. cosec θ
Answer: A. tan θ — Factor sin θ from the numerator and cos θ from the denominator: [sin θ (1 - 2 sin²θ)] / [cos θ (2 cos²θ - 1)]. Since 1 - 2 sin²θ = 1 - 2(1 - cos²θ) = 2 cos²θ - 1, the expression becomes (sin θ/cos θ) * (2 cos²θ - 1) / (2 cos²θ - 1) = tan θ.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.