Important Questions for Class 10 Maths Chapter 8: Introduction to Trigonometry in Class 10 have been compiled to aid students aiming for high scores in their 2023-2024 board exams. These questions, available for download in PDF format on our website, adhere to the latest CBSE syllabus and align with the NCERT book.
Created after a detailed analysis of previous years' question papers, exam trends, and the latest sample papers, these questions are invaluable tools for exam preparation. By solving these questions, students can familiarize themselves with the diverse types of problems that may be presented in the final exam. Students can conveniently learn and solve these problems offline by downloading the Trigonometry Class 10 questions PDF. This resource is designed to provide comprehensive coverage of the trigonometry concept introduced in Chapter 8, elucidating the relationships between angles and sides of a triangle. The questions span various difficulty levels and are accompanied by complete explanations, ensuring a thorough understanding of trigonometric formulas and their application in solving numerical problems.CBSE Important Questions for Class 10 Maths
The six primary trigonometric ratios—sine, cosine, tangent, secant, cosecant, and cotangent—form the foundation of the trigonometry concept explored in this chapter. The questions not only test the application of these ratios but also help students grasp the underlying principles. Expertly solved by our teachers, these important questions serve as a valuable resource for swift comprehension. Students can access Class 10 Maths Chapter 8 Introduction to Trigonometry Multiple-Choice Questions (MCQs) to further enhance their understanding and practice. For additional practice, a set of supplementary questions on Chapter 8 is also provided, allowing students to reinforce their learning and hone their problem-solving skills. Overall, these important questions are crafted to empower students in their preparation for the final examination, offering a holistic understanding of trigonometry in Class 10 Mathematics.Solution:
tan θ + cot θ = 5 … [Given tan2θ + cot2θ + 2 tan θ cot θ = 25 … [Squaring both sides tan2θ + cot2θ + 2 = 25 ∴ tan2θ + cot2θ = 23 Q.3: If tan (A + B) =√3 and tan (A – B) =1/√3, 0° < A + B ≤ 90°; A > B, find A and B.Solution: Given,
tan (A + B) = √3 As we know, tan 60° = √3 Thus, we can write; ⇒ tan (A + B) = tan 60° ⇒(A + B) = 60° …… (i) Now again given; tan (A – B) = 1/√3 Since, tan 30° = 1/√3 Thus, we can write; ⇒ tan (A – B) = tan 30° ⇒(A – B) = 30° ….. (ii) Adding the equation (i) and (ii), we get; A + B + A – B = 60° + 30° 2A = 90° A= 45° Now, put the value of A in eq. (i) to find the value of B; 45° + B = 60° B = 60° – 45° B = 15° Therefore A = 45° and B = 15° Q.4: If tan α = 3 –√ and tan β = 1 3 √ ,0 < α, β < 90°, find the value of cot (α + β). (2012)Solution:
tan α = 3 –√ = tan 60° …(i) tan β = 1 3 √ = tan 30° …(ii) Solving (i) & (ii), α = 60° and β = 30° ∴ cot (α + β) = cot (60° + 30°) = cot 90° = 0 Q.5: If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.Solution: Given,
tan 2A = cot (A – 18°) As we know by trigonometric identities, tan 2A = cot (90° – 2A) Substituting the above equation in the given equation, we get; ⇒ cot (90° – 2A) = cot (A – 18°) Therefore, ⇒ 90° – 2A = A – 18° ⇒ 108° = 3A A = 108° / 3 Hence, the value of A = 36° Q.6: If sec θ + tan θ = 7, then evaluate sec θ – tan θ. (2017OD)Solution:
We know that, sec 2 θ – tan 2 θ = 1 (sec θ + tan θ) (sec θ – tan θ) = 1 (7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7; (Given) ∴ sec θ – tan θ = 1 7 Q.7: Show that : (i) tan 48° tan 23° tan 42° tan 67° = 1 (ii) cos 38° cos 52° – sin 38° sin 52° = 0Solution:
(i) tan 48° tan 23° tan 42° tan 67° We can also write the above given tan functions in terms of cot functions, such as; tan 48° = tan (90° – 42°) = cot 42° tan 23° = tan (90° – 67°) = cot 67° Hence, substituting these values, we get = cot 42° cot 67° tan 42° tan 67° = (cot 42° tan 42°) (cot 67° tan 67°) = 1 × 1 [since cot A.tan A = 1] = 1 (ii) cos 38° cos 52° – sin 38° sin 52° We can also write the given cos functions in terms of sin functions. cos 38° = cos (90° – 52°) = sin 52° cos 52°= cos (90° – 38°) = sin 38° Hence, putting these values in the given equation, we get; sin 52° sin 38° – sin 38° sin 52° = 0 Q.8: If cosec θ = 5 4 , find the value of cot θ. (2014)Solution:
We know that, cot 2 θ = cosec 2 θ – 1 = ( 5 4 ) 2 – 1 ⇒ 25 16 – 1 ⇒ 25 − 16 16 coť 2 θ = 9 16 i cot θ = 3 4 Q.9: If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.Solution:
As we know, for any given triangle, the sum of all its interior angles is equals to 180°. Thus, A + B + C = 180° ….(1) Now we can write the above equation as; ⇒ B + C = 180° – A Dividing by 2 on both the sides; ⇒ (B + C)/2 = (180° – A)/2 ⇒ (B + C)/2 = 90° – A/2 Now, put sin function on both sides. ⇒ sin (B + C)/2 = sin (90° – A/2) Since, sin (90° – A/2) = cos A/2 Therefore, sin (B + C)/2 = cos A/2 Q.10: What happens to value of cos when increases from 0° to 90°? (2015)Solution:
cos 0° = 1, cos 90° = 0 When θ increases from 0° to 90°, the value of cos θ decreases from 1 to 0. Q.11: If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.Solution:
Given, sin θ + cos θ = √3 Squaring on both sides, (sin θ + cos θ) 2 = (√3) 2 sin 2 θ + cos 2 θ + 2 sin θ cos θ = 3 Using the identity sin 2 A + cos 2 A = 1, 1 + 2 sin θ cos θ = 3 2 sin θ cos θ = 3 – 1 2 sin θ cos θ = 2 sin θ cos θ = 1 sin θ cos θ = sin 2 θ + cos 2 θ ⇒ (sin 2 θ + cos 2 θ)/(sin θ cos θ) = 1 ⇒ [sin 2 θ/(sin θ cos θ)] + [cos 2 θ/(sin θ cos θ)] = 1 ⇒ (sin θ/cos θ) + (cos θ/sin θ) = 1 ⇒ tan θ + cot θ = 1 Hence proved. Q.12: Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.Solution:
cot 85° + cos 75° = cot (90° – 5°) + cos (90° – 15°) We know that cos(90° – A) = sin A and cot(90° – A) = tan A = tan 5° + sin 15° Q.13: What is the value of (cos 2 67° – sin 2 23°)?Solution:
(cos 2 67° – sin 2 23°) = cos 2 (90° – 23°) – sin 2 23° We know that cos(90° – A) = sin A = sin 2 23° – sin 2 23° = 0 Therefore, (cos 2 67° – sin 2 23°) = 1. Q.14: Express cot 75° + cosec 75° in terms of trigonometric ratios of angles between 0° and 30°. (2013)Solution:
cot 75° + cosec 75° = cot(90° – 15°) + cosec(90° – 15°) = tan 15° + sec 15° …[cot(90°-A) = tan A cosec(90° – A) = sec A Q.15: Prove that (sin A – 2 sin 3 A)/(2 cos 3 A – cos A) = tan A.Solution:
LHS = (sin A – 2 sin 3 A)/(2 cos 3 A – cos A) = [sin A(1 – 2 sin 2 A)]/ [cos A(2 cos 2 A – 1] Using the identity sin 2 θ + cos 2 θ = 1, = [sin A(sin 2 A + cos 2 A – 2 sin 2 A)]/ [cos A(2 cos 2 A – sin 2 A – cos 2 A] = [sin A(cos 2 A – sin 2 A)]/ [cos A(cos 2 A – sin 2 A)] = sin A/cos A = tan A = RHS Hence proved.Conceptual Reinforcement:
Strategic Preparation:
Diverse Question Types:
Expert Solutions:
Comprehensive Exam Readiness:
Important Questions for Class 10 Maths Chapter 8 |
Chapter 1 Real Numbers |
Chapter 2 Polynomials |
Chapter 3 Linear Equations In Two Variables |
Chapter 4 Quadratic Equations |
Chapter 5 Arithmetic Progression |
Chapter 6 Triangles |
Chapter 7 Coordinate Geometry |
Chapter 8 Introduction to Trigonometry |
Chapter 9 Applications of Trigonometry |
Chapter 10 Circles |
Chapter 11 Constructions |
Chapter 12 Areas Related to Circles |
Chapter 13 Surface Areas and Volumes |
Chapter 14 Statistics |
Chapter 15 Probability |
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