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introduCtion to · PDF file · 2009-09-05Chapter-wise questions 22. In the figure,...
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CHAPTER-WISE QUESTIONS INTRODUCTION TO TRIGONOMETRY 1. Prove cosecθ cotθ – cosecθ cotθ sec (90–θ) cos (90–θ) + tan(90–θ) cosecθ cotθ cot (90–θ) 2. If sin (4A+B) = 1 and tan (2A+B) = √3 , find A and B 3. Given that tan θ = 5 12 , find cos θ and sin θ 4. Prove that 9 sin²θ – 18 + 234 26 cosec²θ = 9 cot²θ cos²θ 5. Evaluate 5 7 sec² 49 – 15 21 cot41 tan 49+ 10 14 tan 82 x tan 43 cot 82 tan 47 6. Prove 2sin²θ – 1 sinθ cosθ = tanθ – cot θ 7. Prove the following : i) 1 (sec x – tanx) – 1 cos x = 1 cos x – 1 (sec x + tanx) ii) (sinθ + secθ)² + (cosθ + cosecθ)² = (1+secθ cosecθ)² iii) (1+ cos θ) sin θ + sin θ (1+ cos θ) = 2 cosecθ iv) (sinθ + 1–cosθ) (cos θ – 1+sinθ) = (1+sinθ) cos θ v) (tanθ + secθ–1) (tanθ – secθ+1) = (1+sinθ) cos θ vi) tanθ (1 – cotθ) + cotθ (1 – tanθ) = 1+secθ cosecθ vii) (1+cotθ – cosecθ)(1+tanθ + secθ) = 2 viii) (cosecθ – cotθ)² = (1–cosθ) (1+cosθ) ix) cos A (1–tanA) + sin A (1–cotA) = sin A + cos A
Transcript of introduCtion to · PDF file · 2009-09-05Chapter-wise questions 22. In the figure,...
Chapter-wise questions
introduCtion to trigonometry
1. Provecosecθcotθ–cosecθ cotθ sec (90–θ)cos (90–θ) + tan(90–θ) cosecθ cotθ
cot (90–θ)
2. Ifsin(4A+B)=1andtan(2A+B)=√3,findAandB
3. Giventhattanθ= 512,findcosθandsinθ
4. Provethat9sin²θ–18+23426 cosec²θ=9cot²θcos²θ
5. Evaluate 57 sec²49–
1521cot41tan49+
1014tan82xtan43cot82tan47
6. Prove 2sin²θ – 1sinθ cosθ =tanθ–cotθ
7. Provethefollowing:
i) 1(sec x – tanx)–
1cos x = 1
cos x –1
(sec x + tanx)
ii) (sinθ + secθ)² + (cosθ + cosecθ)² = (1+secθ cosecθ)²
iii) (1+ cos θ)sin θ
+ sin θ(1+ cos θ)
= 2 cosecθ
iv) (sinθ + 1–cosθ)(cos θ – 1+sinθ)
= (1+sinθ)cos θ
v) (tanθ + secθ–1)(tanθ – secθ+1)
= (1+sinθ)cos θ
vi) tanθ(1 – cotθ)
+ cotθ(1 – tanθ)
= 1+secθ cosecθ
vii) (1+cotθ – cosecθ)(1+tanθ + secθ) = 2
viii) (cosecθ –cotθ)² = (1–cosθ)(1+cosθ)
ix) cos A(1–tanA)
+ sin A(1–cotA)
= sin A + cos A
Chapter-wise questions
x) (1–sinθ + cosθ)² = 2(1+cosθ)(1–sinθ)
xi) √(1+sinθ)(1–sinθ)
+ √(1–sinθ)(1+sinθ)
= 2secθ
8. Ifsecθ+tanθ = m and secθ–tanθ=n,findthevalueof√mn.
9. Ifsinθ+cosθ = p and secθ + cosecθ=q,showthatq(p²–10)=2p.
10.Ifacosθ –bsinθ = x and asinθ + bcosθ = y, prove that a² + b² = x² + y²
11. If x = psecθ+qtanθandy=ptanθ + qsecθ,provethatx²–y² = p²–q²
12. If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2sinθ.
13. If 3tanθ = 4, find the value of: (5sinθ – 3sinθ)(5sinθ + 2cosθ)
14. Without using trigonometric tables evaluate:
i) cos (40°+θ) – sin(50° – θ) = (cos²40° + cos²50°)(sin²40° + sin²50°)
ii) {-tanθ cot(90–θ) + secθ cosec(90–θ) + (sin² 75° +sin²50°)}tan20° tan45° tan50° tan70°
15. If tan A = 512 , find the value of (Sin A + Cos A) Sec A.
16. If sec 4A = cosec (A–20), where 4A is an acute angle, find the value of A
17. If tan A = 43 , find sec A
18. If cot θ = 43 and θ + d = 90º, find the value of tan d.
19. Simplify (1+cot²θ) (1–cosθ)(1+cosθ)
20. Evaluate 2 tan 30º1 – tan²30
21. Find the value of 3 cos 57ºsin 33º – tan 40º
cot 50º
Chapter-wise questions
22. In the figure, AD = 4cm, A
BC
D
θ
12
3
4
BD = 3cm, and CB = 12cm,
find sinθ
23. If cot θ = 34 , find tan²θ
24. Evaluate sin 30º + cos 60º
25. Show that tan 62ºcot 28º = 1
26. Find tan 25º tan 65º – cos 90º
27. In a ∆ABC, right angled at C, if tan A = 1√3
, find the value of
sin A cos B + cos A sinB
28. Without using trigonometric tables, evaluate the following :
sin 18ºcos 72º + √3 (tan 10º tan 30º tan 40º tan 50º tan 80º)
29. Show that cosec²θ – sec²θsecθ cosec²θ +cosecθ sec²θ = cosθ – sin θ
30. If A, B, C are three interior angles of a triangle ABC, show that
cos (B+C2 ) = sin A
2
31. cos (40–θ) – sin (50+θ) + cos²40+cos²50sin²40 + sin²50
32. tanθ = 74 find
(1+sinθ) (1–sinθ)(1+cosθ)(1–cosθ)
33. If 9 sin θ = 12 cos θ, find the value of 3 sinθ + 4 cos θ4 sinθ + 3 cos θ
Chapter-wise questions
34. Prove that cot θ + tan θ = cosecθ sec θ
35. Prove that secθ – 1secθ + 1
= 1–cosθ1+cosθ
36. Prove that 11+sinA
+ 11–sinA
= 2 sec²A
37. Prove that cos²θ (1+tan²θ) = 1
38. Prove that cot A – cos Acot A+cosA = cosec A–1
cosec A+1
39. Prove that (1+cotA–cosecA)(1+tanA+secA) = 2
40. Evaluate without using trigonometric tables
tan 7º. tan 23º. tan 60º. tan 67º. tan 83º + cot 54ºtan 36º + sin20º.sec70º–2
41. Prove that :
sinθ + cosθsinθ – cosθ
+ sinθ – cosθsinθ + cosθ
= 2sec²θtan²θ – 1
42. Evaluate without using Trigonometric Tables:
sec²(90º–θ) – cot²θ(sin²25º + sin²65º)
+ 2cos²60º tan² 28º tan²62º3(sec²43º – cot²47º)
43. If secθ + tanθ = p prove that sinθ = p²–1p²+1
44. Prove 2 sin²θ – 1sinθ.cosθ
= tanθ – cotθ
45. Prove (sinθ + cosecθ)² + (cosθ + secθ )² = 7 + tan²θ + cot²θ
46. Prove cotθ + cosecθ – 11 + cotθ – cosec θ
= sinθ1 – cos θ
47. Prove 1 – sinθ1 + sinθ
= (secθ – tanθ)²
Chapter-wise questions
48. Evaluate sec(42º + θ ) – cosec (48º – θ) + sin 51 cos 39+cos 51 sin 39 –
1cos² 51º
tan²51º49. Prove cosθ
1 – sinθ + cosθ
1 + sinθ = 2 secθ
50. (sinθ + secθ)² + (cosθ + cosecθ)² = (1+secθ cosecθ)²
51. If tanθ + sinθ = m and tanθ – sinθ = n, show that m²–n² = 4√mn
52. Prove that sin θ1 + cosθ
+ 1 + cos θsin θ
= 2 cosec θ
53. sinθ – 2 sin³θ2cos³θ – cosθ
= tanθ
54. (1 + cotA + tanA)(sinA – cosA) = sinA tanA – cotA cosA
55. cos A1–tanA
+ sin A1–cotA
= cos A + sin A
56. (cosecθ – sinθ)(secθ–cosθ) = 1(tanθ+cotθ)
57. sinθcotθ + cosecθ
= 2 + sinθcotθ – cosecθ
58. Prove that √sec²θ + cosec²θ = tanθ + cotθ
59. Prove that tan A1 – cot A
+ cot A1 – tan A
= 1 + tan A + cot A = 1 + secA cosecA
60. Prove that (1+sinθ)²+(1–sinθ)²cos² θ
= 2(1+sin²θ)(1–sin² θ)
61. Prove that cos²θ1 – tanθ
+ sin³θ(sinθ – cosθ)
= 1 + sinθ cosθ
62. Prove that (1+ 1tan²θ
)(1+ 1cot²θ
) = 1sin²θ – sin4θ
63. cos acosb = m and cos a
cosb =nshowthat(m²+n²)cos²b = n²
Chapter-wise questions
64.Provethat:tanθ
1–cotθ + cotθ
1–tanθ =1+tanθ+cotθ
65. Without using trigonometric tables, evaluate the following:
3tan25°tan 40° tan50° tan65° – 12 tan²60°
4(cos²29° + cos²61°)
66.Provethat: 1+cosecθ–cotθ1+cosecθ+cotθ
= cosecθ–cotθ
67.Evaluatethefollowing: cot5°cot10°cot15°cot60°cot75°cot80°cot85°cos² 20° – cos² 70°+2
