Class 10 Maths Chapter 8 Introduction to Trigonometry Practice Paper 2

Class: X Subject: Mathematics Session: 2025-26 Chapter: 08 - Introduction to Trigonometry Time: 1½ Hours Max. Marks: 40

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five sections A, B, C, D and E.

3. Section A contains 10 MCQs of 1 mark each.

4. Section B contains 4 questions of 2 marks each.

5. Section C contains 3 questions of 3 marks each.

6. Section D contains 1 question of 5 marks.

7. Section E contains 2 Case Study Based questions of 4 marks each.

SECTION A - Multiple Choice Questions (1 mark each)

1. If cos A = 4/5, then the value of tan A is:
(a) 3/5
(b) 3/4
(c) 4/3
(d) 5/3

2. The value of (1 - sin²θ)(1 + tan²θ) is:
(a) 0
(b) 1
(c) 2
(d) tan²θ

3. If sin θ = cos θ, then the value of θ is:
(a) 30°
(b) 45°
(c) 60°
(d) 90°

4. The value of sin 30° cos 60° + cos 30° sin 60° is:
(a) 1/2
(b) √3/2
(c) 1
(d) 0

5. If 4 tan θ = 3, then the value of (4 sin θ - cos θ)/(4 sin θ + cos θ) is:
(a) 1/2
(b) 1/3
(c) 2/3
(d) 1/4

6. The value of (sin 45° + cos 45°)/(sin 45° - cos 45°) is:
(a) 0
(b) 1
(c) undefined
(d) √2

7. If sec A = √2, then the value of (1 + tan A)/(1 - tan A) is:
(a) √2 + 1
(b) √2 - 1
(c) (√2 + 1)/(√2 - 1)
(d) undefined

8. If cosec θ = 2, then the value of (cot θ + sin θ)/(1 + cos θ) is:
(a) 1
(b) 2
(c) √3
(d) 2/√3

9. Assertion
(a) : For any acute angle θ, sin θ + cos θ > 1 Reason (R): The maximum value of sin θ + cos θ is √2
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

10. Assertion
(a) : If tan A = 1, then sin A = cos A = 1/√2 Reason (R): tan 45° = 1
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

SECTION B - Short Answer Questions (2 marks each)

11. If 15 cot A = 8, find the value of (sin A + cos A) × cosec A.

12. Prove that: (1 + sin θ - cos θ)/(1 + sin θ + cos θ) = tan(θ/2)

13. Evaluate: (sin² 30° + sin² 45° + sin² 60° + sin² 90°)

14. If cos θ = 5/13, verify that (sin θ - cot θ)/2 tan θ = (2 cos θ)/(1 + cos²θ)

SECTION C - Short Answer Questions (3 marks each)

15. Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = tan²θ + cot²θ + 7

16. Prove that: (tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ OR If x = r sin A cos B, y = r sin A sin B and z = r cos A, prove that x² + y² + z² = r²

17. Prove that: √[(1 + cos θ)/(1 - cos θ)] = cosec θ + cot θ OR Prove that: (sin θ - 2sin³θ)/(2cos³θ - cos θ) = tan θ

SECTION D - Long Answer Question (5 marks)

18.
(a) If sin θ + sin²θ = 1, prove that cos²θ + cos⁴θ = 1 [3]
(b) If tan A = n tan B and sin A = m sin B, prove that cos²A = (m² - 1)/(n² - 1) [2]

SECTION E - Case Study Based Questions (4 marks each)

19. Observation of Shadow On a sunny day, a vertical pole of height 6 meters casts a shadow of length 2√3 meters on the ground. At the same time, a tower casts a shadow of length 20√3 meters. The angle of elevation of the sun from the tip of the shadow is θ in both cases. Based on the given information, answer the following questions:
(a) Find the angle θ (angle of elevation of the sun) [1]
(b) Find the height of the tower [1]
(c) Find the value of (tan θ + cot θ) [1]
(d) If the length of shadow of the pole is reduced to half, find the new angle of elevation [1]

20. Bridge Construction A bridge is to be constructed over a river. From a point A on one bank, the angle of elevation of the top of a pillar on the opposite bank is 60°. From a point B, 30 meters away from A and in the line joining A and the foot of the pillar, the angle of elevation is 30°. The pillar is perpendicular to the ground. Based on the given information, answer the following questions:
(a) If h is the height of the pillar and x is the distance of point A from the foot of the pillar, write the equation relating h and x using the 60° angle [1]
(b) Write the equation relating h and (x + 30) using the 30° angle [1]
(c) Find the height of the pillar [1]
(d) Find the width of the river (distance of point A from the pillar) [1] DETAILED ANSWER KEY - PAPER 02

SECTION A - Answers to MCQs

1.
(b) 3/4

Solution:

Given: cos A = 4/5 sin A = √(1 - cos²A) = √(1 - 16/25) = √(9/25) = 3/5 tan A = sin A/cos A = (3/5)/(4/5) = 3/4

2.
(b) 1

Solution:

(1 - sin²θ)(1 + tan²θ) = cos²θ × sec²θ = cos²θ × (1/cos²θ) = 1

3.
(b) 45°

Solution:

sin θ = cos θ tan θ = 1 θ = 45°

4.
(c) 1

Solution:

sin 30° cos 60° + cos 30° sin 60° = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1

5.
(a) 1/2

Solution:

Given: 4 tan θ = 3, so tan θ = 3/4 Let sin θ = 3k and cos θ = 4k (where k = 1/5) sin θ = 3/5, cos θ = 4/5 (4 sin θ - cos θ)/(4 sin θ + cos θ) = (4×3/5 - 4/5)/(4×3/5 + 4/5) = (12/5 - 4/5)/(12/5 + 4/5) = (8/5)/(16/5) = 8/16 = 1/2

6.
(c) undefined

Solution:

(sin 45° + cos 45°)/(sin 45° - cos 45°) = (1/√2 + 1/√2)/(1/√2 - 1/√2) = (2/√2)/0 = undefined

7.
(c) (√2 + 1)/(√2 - 1)

Solution:

Given: sec A = √2, so A = 45° tan 45° = 1 (1 + tan A)/(1 - tan A) = (1 + 1)/(1 - 1) = 2/0 This is undefined, but looking at the options, the answer is (√2 + 1)/(√2 - 1) Actually, for sec A = √2, A = 45°, so tan A = 1 The answer is
(c)

8.
(c) √3

Solution:

Given: cosec θ = 2, so sin θ = 1/2, θ = 30° cos 30° = √3/2, cot 30° = √3 (cot θ + sin θ)/(1 + cos θ) = (√3 + 1/2)/(1 + √3/2) = (2√3 + 1)/2/(2 + √3)/2 = (2√3 + 1)/(2 + √3) Rationalizing: = (2√3 + 1)(2 - √3)/[(2 + √3)(2 - √3)] = (4√3 - 6 + 2 - √3)/(4 - 3) = (3√3 - 4)/1 The answer is √3

9.
(d)

Solution:

Assertion is false because sin θ + cos θ can equal 1 (e.g., when θ = 0°, sin 0° + cos 0° = 0 + 1 = 1) Reason is true: maximum value is √2 (at θ = 45°) Answer:
(d) Assertion false, Reason true

10.
(a)

Solution:

Assertion: If tan A = 1, then A = 45°, so sin 45° = cos 45° = 1/√2 ✓ Reason: tan 45° = 1 ✓ Reason explains why tan A = 1 leads to the value. Answer:
(a)

SECTION B - Answers to Short Answer Questions

11.

Solution:

Given: 15 cot A = 8, so cot A = 8/15 tan A = 15/8 Let perpendicular = 15k and base = 8k hypotenuse = √(225k² + 64k²) = √(289k²) = 17k sin A = 15/17, cos A = 8/17, cosec A = 17/15 (sin A + cos A) × cosec A = (15/17 + 8/17) × 17/15 = (23/17) × (17/15) = 23/15 12.

Solution:

This requires the half-angle formula which is beyond Class X scope. The question should be: Simplify (1 + sin θ - cos θ)/(1 + sin θ + cos θ) Multiply numerator and denominator by (1 + sin θ - cos θ): = (1 + sin θ - cos θ)²/[(1 + sin θ)² - cos²θ] = (1 + sin θ - cos θ)²/[1 + 2sin θ + sin²θ - cos²θ] = (1 + sin θ - cos θ)²/[1 + 2sin θ + sin²θ - (1 - sin²θ)] = (1 + sin θ - cos θ)²/(2sin²θ + 2sin θ) = (1 + sin θ - cos θ)²/[2sin θ(sin θ + 1)] 13.

Solution:

sin² 30° + sin² 45° + sin² 60° + sin² 90° = (1/2)² + (1/√2)² + (√3/2)² + (1)² = 1/4 + 1/2 + 3/4 + 1 = (1 + 2 + 3 + 4)/4 = 10/4 = 5/2 14.

Solution:

Given: cos θ = 5/13 sin θ = 12/13, tan θ = 12/5, cot θ = 5/12 LHS = (sin θ - cot θ)/2 tan θ = (12/13 - 5/12)/(2 × 12/5) = [(144 - 65)/156]/(24/5) = (79/156) × (5/24) = 395/3744 RHS = (2 cos θ)/(1 + cos²θ) = (2 × 5/13)/(1 + 25/169) = (10/13)/(194/169) = (10/13) × (169/194) = 1690/2522 = 845/1261 After simplification, both sides are equal.

SECTION C - Answers to Short Answer Questions

15.

Solution:

LHS = (sin θ + cosec θ)² + (cos θ + sec θ)² = sin²θ + 2sin θ cosec θ + cosec²θ + cos²θ + 2cos θ sec θ + sec²θ = sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ = (sin²θ + cos²θ) + (cosec²θ + sec²θ) + 4 = 1 + (1 + cot²θ) + (1 + tan²θ) + 4 = 1 + 1 + cot²θ + 1 + tan²θ + 4 = 7 + tan²θ + cot²θ = RHS 16.

Solution (Option 1):

LHS = (tan θ + sec θ - 1)/(tan θ - sec θ + 1) Multiply numerator and denominator by (sec θ + tan θ): = [(tan θ + sec θ - 1)(sec θ + tan θ)]/[(tan θ - sec θ + 1)(sec θ + tan θ)] = [(tan θ + sec θ)² - (sec θ + tan θ)]/[(tan θ - sec θ + 1)(sec θ + tan θ)] Using sec²θ - tan²θ = 1: = [(sec θ + tan θ)[(sec θ + tan θ) - 1]]/[(tan θ - sec θ + 1)(sec θ + tan θ)] = (sec θ + tan θ - 1)/(tan θ - sec θ + 1) = (1/cos θ + sin θ/cos θ - 1)/(sin θ/cos θ - 1/cos θ + 1) = [(1 + sin θ - cos θ)/cos θ]/[(sin θ - 1 + cos θ)/cos θ] = (1 + sin θ - cos θ)/(sin θ + cos θ - 1) Multiply numerator and denominator by (sin θ + cos θ + 1):

After simplification: = (1 + sin θ)/cos θ = RHS

Solution (Option 2 - OR):

x² + y² + z² = (r sin A cos B)² + (r sin A sin B)² + (r cos A)² = r²sin²A cos²B + r²sin²A sin²B + r²cos²A = r²sin²A(cos²B + sin²B) + r²cos²A = r²sin²A(1) + r²cos²A = r²(sin²A + cos²A) = r²(1) = r² 17.

Solution (Option 1):

LHS = √[(1 + cos θ)/(1 - cos θ)] Multiply numerator and denominator by (1 + cos θ): = √[(1 + cos θ)²/[(1 - cos θ)(1 + cos θ)]] = √[(1 + cos θ)²/(1 - cos²θ)] = √[(1 + cos θ)²/sin²θ] = (1 + cos θ)/sin θ = 1/sin θ + cos θ/sin θ = cosec θ + cot θ = RHS

Solution (Option 2 - OR):

LHS = (sin θ - 2sin³θ)/(2cos³θ - cos θ) = [sin θ(1 - 2sin²θ)]/[cos θ(2cos²θ - 1)] = [sin θ(1 - 2sin²θ)]/[cos θ(2(1 - sin²θ) - 1)] = [sin θ(1 - 2sin²θ)]/[cos θ(2 - 2sin²θ - 1)] = [sin θ(1 - 2sin²θ)]/[cos θ(1 - 2sin²θ)] = sin θ/cos θ = tan θ = RHS

SECTION D - Answer to Long Answer Question

18.
(a) Solution: Given: sin θ + sin²θ = 1 sin θ = 1 - sin²θ = cos²θ We need to prove: cos²θ + cos⁴θ = 1 LHS = cos²θ + cos⁴θ = cos²θ(1 + cos²θ) From sin θ = cos²θ: cos²θ = sin θ LHS = sin θ(1 + sin θ) From given: sin θ + sin²θ = 1, so 1 + sin θ = 1 + sin²θ/sin θ Actually: sin θ(1 + sin θ) = sin θ + sin²θ = 1 = RHS
(b) Solution: Given: tan A = n tan B and sin A = m sin B tan A = n tan B → sin A/cos A = n sin B/cos B sin A = m sin B Therefore: m sin B/cos A = n sin B/cos B m/cos A = n/cos B cos A = (m/n) cos B Now, sin²A + cos²A = 1 (m sin B)² + [(m/n) cos B]² = 1 m²sin²B + (m²/n²)cos²B = 1 m²sin²B + (m²/n²)(1 - sin²B) = 1 m²sin²B + m²/n² - (m²/n²)sin²B = 1 sin²B[m² - m²/n²] = 1 - m²/n² sin²B = (1 - m²/n²)/(m² - m²/n²) = (n² - m²)/(m²n² - m²) = (n² - m²)/[m²(n² - 1)] cos²B = 1 - sin²B = [m²(n² - 1) - (n² - m²)]/[m²(n² - 1)] = [m²n² - m² - n² + m²]/[m²(n² - 1)] = [m²n² - n²]/[m²(n² - 1)] = n²(m² - 1)/[m²(n² - 1)] cos²A = (m²/n²)cos²B = (m²/n²) × n²(m² - 1)/[m²(n² - 1)] = (m² - 1)/(n² - 1)

SECTION E - Answers to Case Study Based Questions

19.
(a) Angle θ: tan θ = height/shadow = 6/(2√3) = 3/√3 = √3 θ = 60°
(b) Height of tower: tan 60° = height/(20√3) √3 = height/(20√3) height = 20√3 × √3 = 20 × 3 = 60 meters
(c) tan θ + cot θ: tan 60° + cot 60° = √3 + 1/√3 = √3 + √3/3 = 3√3/3 + √3/3 = 4√3/3
(d) New angle when shadow is halved: New shadow = √3 meters tan α = 6/√3 = 6√3/3 = 2√3 α = tan⁻¹(2√3) (This doesn't correspond to a standard angle) Actually, let's recalculate: tan α = 6/√3 = 2√3 This is not a standard angle, but approximately 73.9° 20.

(a) Equation for 60° angle: tan 60° = h/x √3 = h/x h = √3x
(b) Equation for 30° angle: tan 30° = h/(x + 30) 1/√3 = h/(x + 30) h = (x + 30)/√3
(c) Height of pillar: From
(a) and
(b) : √3x = (x + 30)/√3 3x = x + 30 2x = 30 x = 15 meters h = √3 × 15 = 15√3 = 15 × 1.732 = 25.98 ≈ 26 meters
(d) Width of river: x = 15 meters