Bihar Board Class 10 Model Question Paper 2022 Maths Advanced

SECONDARY SCHOOL EXAMINATION – 2022
(ANNUAL)
Sub. Code –114/214
111142222221111 Model Question Paper
le;% 3 ?kaVs 15 feuV ¼vkn”kZ iz”u i=½
Time: 3 Hours 15 Minutes
Advanced Mathematics (Optional)
mPp xf.kr ¼,sfPNd½

Total no. of Questions : 100+30+8 = 138 Full Marks - 100

Instructions to the Candidates :

1- ijh{kkFkhZ OMR mÙkj i=d ij viuk iz”u iqfLrdk Øekad ¼10 vadksa dk½ vo”;
fy[ksaA
Candidate must enter his/her Question Booklet Serial No. (of 10
digits) in the OMR Answer Sheet.

2- ijh{kkFkhZ ;FkklaHko vius 'kCnksa esa gh mÙkj nsaA
Candidates are required to give their answers in own words as far
as practicable.

3- nkfguh vksj gkf”k, ij fn;s gq, vad iw.kkZad fufnZ"V djrs gSaA
Figures in the right hand margin indicate full marks.

4- iz”uksa dks /;kuiwoZd i<+us ds fy, ijh{kkfFkZ;ksa dks 15 feuV dk vfrfjDr le;
fn;k x;k gSA

1

15 minutes of extra time has been allotted to the candidates to
read the questions carefully.

5- ;g iz”u iqfLrdk nks [k.Mksa esa gS & ,oa A
This question booklet is divided into two sections – Section-A and
Section-B.
6- [k.M&v esa 100 oLrqfu"B iz”u gSa] ftuesa ls fdUgha 50 iz”uksa dk mÙkj nsuk
vfuok;Z gSA 50 ls vf/kd iz”uksa ds mÙkj nsus ij izFke 50 dk gh ewY;kadu
dEI;wVj }kjk fd;k tk,xkA izR;sd iz”u ds fy, 1 vad fu/kkZfjr gSA lgh mÙkj
dks miyC/k djk, x;s OMR mÙkj i=d esa fn, x, lgh fodYi dks
uhys@dkys ckWy isu ls izxk<+ djsaA fdlh Hkh izdkj ds g~okbVuj @ rjy
inkFkZ @ CysM @ uk[kwu vkfn dk OMR mÙkj i=d esa iz;ksx djuk euk gS]
vU;Fkk ifj.kke vekU; gksxkA
In Section-A, there are 100 objective type questions, out of which
any 50 questions are to be answered. First 50 answers will be
evaluated by the computer in case more than 50 questions are
answered. Each question carries 1 mark. Darken the circle with
blue / black ball pen against the correct option on OMR Answer
Sheet provided to you. Do not use Whitener / liquid / blade / nail
etc. on OMR-sheet, otherwise the result will be invalid.

7- [k.M&c esa 30 y?kq mÙkjh; iz”u gSa] ftuesa ls fdUgha 15 iz”uksa dk mÙkj nsuk
vfuok;Z gSA izR;sd iz”u ds fy, 2 vad fu/kkZfjr gSA buds vfrfjDr] bl [k.M
esa 8 nh?kZ mÙkjh; iz”u fn;s x;s gSa] ftuesa ls fdUgha 4 iz”uksa dk mÙkj nsuk
vfuok;Z gSA izR;sd ds fy, 5 vad fu/kkZfjr gSA

2

In Section-B, there are 30 short answer type questions, out of
which any 15 questions are to be answered. Each question carries
2 marks. Apart from these, there are 8 long answer type questions,
out of which any 4 questions are to be answered. Each question
carries 5 marks.

8- fdlh izdkj ds bysDVªkWfud midj.k dk iz;ksx iw.kZr;k oftZr gSA
Use of any electronic appliances is strictly prohibited.

3

[k.M & v @ Section - A
oLrqfu"B iz”u @ Objective Type Qestions

iz”u la[;k 1 ls 100 rd ds iz”u ds lkFk pkj fodYi fn, x, gSa ftuesa ls ,d lgh gSA
fdUgha 50 iz”uksa ds mÙkj nsaA vius }kjk pqus x, lgh fodYi dks OMR 'khV ij fpfUgr
djsaA 50x1=50
Question nos. 1 to 100 have four options, out of which only one is correct.
Answer any 50 questions. You have to mark your selected option on the
OMR-sheet. 50x1=50

1. dks.k 800 dk eku jsfM;u esa gS

(A) 4𝜋 (B) 𝜋

(C) 𝜋 (D) 𝜋

The value of the angle 800 in radian is

(A) 4𝜋 (B) 𝜋

(C) 𝜋 (D) 𝜋

2. 300 xzsM fdrus ledks.k ds cjkcj gS \

(A) 3 (B) 4

(C) 2 (D) 5

300 grade is equal to how many right angles ?

(A) 3 (B) 4

(C) 2 (D) 5

4

3. fcUnq (-5, -4) fLFkr gS

(A) izFke ikn esa (B) f}rh; ikn esa

(C) r`rh; ikn esa (D) prqFkZ ikn esa

Point (-5, -4) lies in

(A) first quadrant (B) second quadrant

(C) third quadrant (D) fourth quadrant

4. 2790 fdl ikn esa fLFkr gS \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

In which quadrant is 2790 situated ?

(A) First (B) Second

(C) Third (D) Fourth

5. fcUnq (6, 0) dh dksfV gS

(A) 6 (B) 0

(C) 5 (D) 12

Ordinate of the point (6, 0) is

(A) 6 (B) 0

(C) 5 (D) 12

6. cos1350 dk eku gksxk

5

(A) (B) -
√ √

(C) (D) √2

The value of cos1350 is

(A) (B) -
√ √

(C) (D) √2

7. ¾

(A) 0 (B) 1

(C) 2 (D) -1

8. Sin(900 - 𝜃) =

(A) cos𝜃 (B) sin𝜃

(C) tan𝜃 (D) cosec𝜃

9. sin(A + B) =

(A) sinA.cosB – cosA.sinB (B) cosA.sinB - sinA.cosB

(C) sinA.cosB + cosA.sinB (D) cosA.cosB – sinA.sinB

10. cos(A - B) =

(A) cosA.cosB + sinA.sinB (B) cosA.cosB – sinA.sinB

(C) cosA.sinB + sinA.cosB (D) sinA.cosB – sinB.cosA

11. sin(A + B).sin(A – B) =

(A) cos2B – cos2A (B) cos2A – cos2B
6

(C) cos2B – sin2A (D) cos2A – sin2B

12. sin𝜃.tan𝜃 =

(A) (B)

(C) (D) cos𝜃

13. cosec(900 + 𝜃) =

(A) cos𝜃 (B) sec𝜃

(C) -cosec𝜃 (D) tan𝜃

14. ;fn sin𝜃 = gks] rks sec𝜃 dk eku gksxk

(A) (B)

(C) (D)

If sin𝜃 = , then the value of sec𝜃 is

(A) (B)

(C) (D)

15. 2cosA.cosB =

(A) cos(A + B) + cos(A – B) (B) 2cosA.sinB

(C) cos(A – B) – cos(A + B) (D) 2sinA.cosB

16. sin2A =

(A) (B) cos2A – sin2A
7

(C) 2cos2A (D) 2sin2A

17. fcUnq (0, 8) dh Hkqt gksxh

(A) 0 (B) 8

(C) (D)

The abscissa of the point (0, 8) is

(A) 0 (B) 8

(C) (D)

18. ;fn sin𝜃 = √ gks] rks 𝜃 dk eku gksxk

(A) 300 (B) 450

(C) 600 (D) 900

If sin𝜃 = , then the value of 𝜃 will be
√

(A) 300 (B) 450

(C) 600 (D) 900

19. ;fn r = 10 lseh] l = 20 lseh gks] rks 𝜃 ¼jsfM;u esa½ dk eku gS

(A) 4 (B) 2

(C) 1 (D) 3

If r = 10 cm, l = 20 cm, then the value of 𝜃 (in radian) is

(A) 4 (B) 2

(C) 1 (D) 3

8

20. ;fn sin𝜃 = gks] rks cos𝜃 dk eku gS

(A) (B)

(C) (D)

If sin𝜃 = , then the value of cos𝜃 is

(A) (B)

(C) (D)

21. ewy fcUnq ls fcUnq ¼5] 12½ dh nwjh gS

(A) 13 (B) 17

(C) 7 (D) 30

The distance of the point (5, 12) from the origin is

(A) 13 (B) 17

(C) 7 (D) 30

22. ;fn cosecA = √2 gks] rks cot2A dk eku gksxk

(A) 1 (B) 2

(C) √2 (D) 3

If cosecA = √2, then the value of cot2A will be

(A) 1 (B) 2

(C) √2 (D) 3

9

23. Sin2450 + cos2450 =

(A) 0 (B) 1

(C) -1 (D) 2

24. ;fn A = B = 450 gks] rks sin(A – B) =

(A) 1 (B) 0

(C) -1 (D) 2

If A = B = 450 , then sin(A – B) =

(A) 1 (B) 0

(C) -1 (D) 2

25. ;fn 3sin𝜃 = 2 gks] rks cosec𝜃 dk eku gksxk

(A) (B)

(C) (D)

If 3sin𝜃 = 2, then the value of cosec𝜃 will be

(A) (B)

(C) (D)

26. fcUnqvksa (7cos𝜃, 0) rFkk (0, 7sin𝜃) ds chp dh nwjh gS

(A) 6 (B) 7

(C) 14 (D) 49

Distance between the points (7cos𝜃, 0) and (0, 7sin𝜃) is

10

(A) 6 (B) 7

(C) 14 (D) 49

27. fuEufyf[kr esa dkSu 7 ds cjkcj gS \

(A) 7sec2𝜃 – 7tan2𝜃 (B) 7sec2𝜃 + 7 tan2𝜃

(C) 1 – 7cos2𝜃 (D) 7sec2𝜃 – 7cos2𝜃

Which of the following is equal to 7 ?

(A) 7sec2𝜃 – 7tan2𝜃 (B) 7sec2𝜃 + 7 tan2𝜃

(C) 1 – 7cos2𝜃 (D) 7sec2𝜃 – 7cos2𝜃

28. tan2600 – tan2450 =

(A) 2 (B) 1

(C) -1 (D) 0

29. √3tan600 + tan450 =

(A) 4 (B) 3

(C) 2 (D) 1

30. ;fn sin𝜃 = , tgka 𝜃 ¼dks.k½ ] 0 rFkk 𝜋 ds chp esa gks] rks 𝜃 dk eku gksxk

(A) 600,1200 (B) 300, 1500

(C) 300, 1350 (D) 300, 1200

If sin𝜃 = , where angle 𝜃 lies between 0 and 𝜋, then the value of

𝜃 will be

11

(A) 600,1200 (B) 300, 1500

(C) 300, 1350 (D) 300, 1200

31. tan (450 – A) =

(A) (B)

(C) 1 + tanA (D) 1 - tanA

32. ;fn sin𝜃 = 0.1 gks] rks sin3𝜃 dk eku gS

(A) 0.286 (B) 0.386

(C) 0.296 (D) 0.396

If sin𝜃 = 0.1, then the value of sin3𝜃 will be

(A) 0.286 (B) 0.386

(C) 0.296 (D) 0.396

33. ;fn 2𝜃 = 900 gks] rks tan𝜃 dk eku gS

(A) 1 (B) −1

(C) √3 (D)
√

If 2𝜃 = 900, then the value of tan𝜃 will be

(A) 1 (B) −1

(C) √3 (D)
√

34. 2sinA.cosB =

(A) sin(A + B) + sin (A – B) (B) sin(A + B) – sin (A – B)

12

( )
(C) sin(A + B).sin(A – B) (D)
( )

35. 1 – cos2600 =

(A) (B)

√
(C) (D)

36. =

(A) tan2A (B) cos2A

(C) sin2A (D) sec2A

37. Sin3A + 4sin3A =

(A) 3sinA (B) 3cosA

(C) 3cosecA (D) 3secA

38. =

(A) cos3A (B) cos2A

(C) sin2A (D) cot2A

39. f=Hkqt ftlds 'kh"kZ (1, 2), (4, 7) rFkk (7, -3) gaS] rks dsUnzd ds funsZ”kkad gS

(A) (4, 2) (B) (2, 4)

(C) (4, 1) (D) buesa ls dksbZ ugha

The coordinates of the centroid of the triangle whose vertices are

(1, 2), (4, 7) and (7, -3) are

(A) (4, 2) (B) (2, 4)

13

(C) (4, 1) (D) None of these

40. y &v{k ls fcUnq ¼3] 7½ dh nwjh gS

(A) 3 (B) 9

(C) 11 (D) 7

The distance of the point (3, 7) from the y-axis is

(A) 3 (B) 9

(C) 11 (D) 7

41. fcUnq ¼0] 0½ vkSj ¼6] 10½ dks feykusokyh js[kk[k.M ds e/; fcUnq ds fu;ked ga S

(A) (-6, -10) (B) (6, 5)

(C) (3, 5) (D) (-3, -5)

The coordinates of mid point of the line segment joining the points

(0, 0) and (6, 10) are

(A) (-6, -10) (B) (6, 5)

(C) (3, 5) (D) (-3, -5)

42. 4cos𝜃 + 3sin𝜃 dk eku 𝜃= 900 ij gS

(A) 7 (B) 6

(C) 3 (D) buesa ls dksbZ ugha

The value of 4cos𝜃 + 3sin𝜃 at 𝜃= 900 is

(A) 7 (B) 6

(C) 3 (D) None of these

14

43. fcUnqvksa (P, Q) rFkk (-P, -Q) ds chp dh nwjh gS

(A) 𝑃 + 𝑄 (B) 2 𝑃 + 𝑄

(C) 2(𝑃 + 𝑄 ) (D) 1

Distance between the points (P, Q) and (-P, -Q) is

(A) 𝑃 + 𝑄 (B) 2 𝑃 + 𝑄

(C) 2(𝑃 + 𝑄 ) (D) 1

44. fdlh o`Ùk ds O;kl ds Nksjksa ds funsZ”kkad ¼4] &3½ rFkk ¼&4] 3½ gaS] rks o`Ùk ds

dsUnz ds funZ”kkad gSa

(A) (4, 3) (B) (-4, 3)

(C) (0, 0) (D) (-4, -3)

If (4, -3) and (-4, 3) are the coordinates of the ends of the diameter

of the circle, then the coordinates of centre of the circle are

(A) (4, 3) (B) (-4, 3)

(C) (0, 0) (D) (-4, -3)

45. cos(900 - 𝜃) =

(A) sin𝜃 (B) -sin𝜃

(C) cos𝜃 (D) -cos𝜃

46. tan150 =

(A) 2 - √3 (B) 2 + √3

15

(C) √3 (D) 2

47. sinC – sinD =

(A) 2cos .sin (B) 2sin .cos

(C) cos .sin (D) sin .cos

48. =

(A) cos2A (B) sin2A

(C) cosA (D) sinA

49. cosA + sin2 =

(A) cos2 (B) sin2

(C) cos2A (D) sin2A

50. 1 + cosA =

(A) 2cos (B) 2cos2

(C) cos2 (D) cos

51. Sin𝜃 =

(A) √1 − 𝑐𝑜𝑠2𝜃 (B)

(C) (D) √1 + 𝑐𝑜𝑠2𝜃

52. ;fn A + B + C = 𝜋 gks] rks tan(A + B) =

16

(A) tanA (B) tanB

(C) tanC (D) –tanC

If A + B + C = 𝜋, then tan(A + B) =

(A) tanA (B) tanB

(C) tanC (D) –tanC

53. fdlh dks.k ds 1500 dh fLFkfr esa ifjHkze.k fdj.k dh fLFkfr gksxh

(A) izFke ikn esa (B) f}rh; ikn esa

(C) r`rh; ikn esa (D) prqFkZ ikn esa

The position of rotating ray in the case of 1500 will be

(A) in 1st quadrant (B) in 2nd quadrant

(C) in 3rd quadrant (D) in 4th quadrant

54. &6000 fdl ikn esa fLFkr gS \

(A) izFke ikn esa (B) f}rh; ikn esa

(C) r`rh; ikn esa (D) prqFkZ ikn esa

In which quadrant is -6000 situated

(A) in 1st quadrant (B) in 2nd quadrant

(C) in 3rd quadrant (D) in 4th quadrant

55. fdl ikn esa fdlh fcUnq ds Hkqt /kukRed vkSj dksfV _.kkRed gksaxs \

(A) izFke ikn esa (B) f}rh; ikn esa

(C) r`rh; ikn esa (D) prqFkZ ikn esa

17

In which quadrant a point will have its abscissa positive and

ordinate negative ?

(A) in 1st quadrant (B) in 2nd quadrant

(C) in 3rd quadrant (D) in 4th quadrant

56. 3 cts ?kM+h dh feuV dh lqbZ vkSj ?kaVs dh lqbZ ds chp dk dks.k gksxk

(A) 900 (B) 600

(C) 300 (D) 1800

The angle between the minute and hour hands of a clock at 3

O’clock will be

(A) 900 (B) 600

(C) 300 (D) 1800

57. sec𝜃 dk eku cos𝜃 ds inksa esa gksxk

(A) cos𝜃 (B)

(C) 2cos𝜃 (D) cos2𝜃

The value of sec𝜃 in terms of cos𝜃 will be

(A) cos𝜃 (B)

(C) 2cos𝜃 (D) cos2𝜃

58. fdlh f=Hkqt ds dks.k 3%4%5 ds vuqikr esa gS] rks lcls NksVs dsk.k dk eku gksxk

(A) 600 (B) 450

18

(C) 750 (D) 300

The angles of a triangle are in the ratio 3:4:5 then the value of the

smallest angle will be

(A) 600 (B) 450

(C) 750 (D) 300

59. ?kM+h esa lsds.M okyh lqbZ 2700 dk dks.k fdrus lsds.M esa cukrh gS \

(A) 30 lsds.M (B) 45 lsds.M

(C) 60 lsds.M (D) 40 lsds.M

How much times does the second hand of a clock take to

complete an angle equal to 2700?

(A) 30 seconds (B) 45 seconds

(C) 60 seconds (D) 40 seconds

60. ;fn tan𝜃 = tan450 gks] rks 𝜃 dk eku gS

(A) 300 (B) 450

(C) 600 (D) 900

If tan𝜃 = tan450, then the value of 𝜃 is

(A) 300 (B) 450

(C) 600 (D) 900

61. tan600.tan300 dk eku gS

(A) 1 (B)

19

(C) (D)
√

The value of tan600.tan300 is

(A) 1 (B)

(C) (D)
√

62. tan300 + tan1500 =

(A) 1 (B) 0

(C) √3 (D)
√

63. ;fn tan𝛼 = 𝑃 gks] rks cos2𝛼 cjkcj gksxk

(A) (B)

(C) (D)

If tan𝛼 = 𝑃, then cos2𝛼 is equal to

(A) (B)

(C) (D)

64. ;fn A = 600 gks] rks cosA dk eku gksxk

(A) (B)
√

√
(C) (D)
√

If A = 600, then the value of cosA is

20

(A) (B)
√

√
(C) (D)
√

65. 2sin750.cos150 =

√ √
(A) (B)

√
(C) (D) 2 + √3

66. =

(A) (B) √3
√

(C) 1 (D) 2

67. 9sin – 12𝑠𝑖𝑛 =

(A) 3 (B) 4

(C) 2 (D) 1

68. In ∆ABC, tan .tan =

(A) (B)

(C) (D)

In ∆ABC, tan .tan =

(A) (B)

21

(C) (D)

69. In ∆ABC, B = 300, b=6, a=8] rFkk sinA=x gks] rks x=

(A) (B)

(C) (D)

In ∆ABC, B = 300, b=6, a=8 and sinA=x then x=

(A) (B)

(C) (D)

70. ;fn fdlh ∆ABC esa a, b rFkk C fn;s x;s gkas] rks ∆ABC dk {ks=Qy gS

(A) absinC (B) abcosC

(C) absinC (D) 2absinC

If in ∆ABC, a, b and C are given, then the area of ∆ABC is

(A) absinC (B) abcosC

(C) absinC (D) 2absinC

71. ∆ABC esa b2 + 2cacosB =

(A) c2 – a2 (B) c2 + a2

(C) c + a (D) c – a

In ∆ABC, b2 + 2cacosB =

(A) c2 – a2 (B) c2 + a2

22

(C) c + a (D) c – a

72. ∆ABC esa tan =

(A) cot (B) cot

(C) cot (D) cot2

In ∆ABC, tan =

(A) cot (B) cot

(C) cot (D) cot2

73. sin400 - sin200 dk eku gksxk

(A) √3sin100 (B) sin100

(C) √3 (D) √3sin200

The value of sin400 - sin200 is

(A) √3sin100 (B) sin100

(C) √3 (D) √3sin200

74. tan750 =

(A) 2 - √3 (B) 2 + √3

√ √
(C) (D)
√ √

75. cos =

23

(A) 1 (B) -1

(C) (D) -
√ √

76. cot(𝜋+𝜃) =

(A) sin𝜃 (B) tan𝜃

(C) cot𝜃 (D) co𝜃

77. ;fn cosx = gks] rks cos3x = ?

(A) (B)

(C) (D)

If cosx = , then cos3x = ?

(A) (B)

(C) (D)

78. fuEufyf[kr easa dkSu fcUnq prqFkZ ikn esa gS \

(A) (4, 7) (B) (6, -7)

(C) (-6, 7) (D) (-9, -6)

Which of the following points lies in fourth quadrant ?

(A) (4, 7) (B) (6, -7)

(C) (-6, 7) (D) (-9, -6)

79. ;fn 𝜃 = 900 gks] rks sin – cos dk eku gksxk

24

(A) 1 (B) -1

(C) 2 (D) 0

If 𝜃 = 900, then the value of sin – cos is

(A) 1 (B) -1

(C) 2 (D) 0

80. ;fn √3tan𝜃 = 1 gks] rks 𝜃 dk eku gksxk

(A) 300 (B) 600

(C) 450 (D) 900

If √3tan𝜃 = 1, then the value of 𝜃 is

(A) 300 (B) 600

(C) 450 (D) 900

81. 1 + cot2𝜃 =

(A) sec2𝜃 (B) cosec2𝜃

(C) tan2𝜃 (D) sin2𝜃

82. Sin400.cos500 + cos400.sin500 =

(A) 1 (B) -1

(C) 2 (D) -2

83. 𝜃 ds fdl eku ds fy, fuEufyf[kr esa dkSu vlaHko gS \

(A) sin𝜃 = (B) sec𝜃 = √3
√

25

√
(C) tan𝜃 = (D) sin𝜃 =

Which of the following is impossible for a value of 𝜃 ?

(A) sin𝜃 = (B) sec𝜃 = √3
√

√
(C) tan𝜃 = (D) sin𝜃 =

84. Cos1050 =

√ √
(A) (B)
√

√ √
(C) (D)
√ √

85. 2sin5A.cos5A =

(A) sin5A (B) sin10A

(C) sin15A (D) sin25A

86. =
√

(A) 2 (B) 1

(C) -1 (D) -2

87. 2sin450.cos450 =

(A) sin450 (B) sin900

(C) sin600 (D) sin300

88. ;fn cosA = vkSj cosB = gks] rks tan(A-B) dk eku gS

26

(A) (B)

(C) (D) a

If cosA = and cosB = , then the value of tan(A-B) is

(A) (B)

(C) (D)

89. cos𝜃.tan𝜃 =

(A) sin𝜃 (B) cos𝜃

(C) tan𝜃 (D) cot𝜃

90. sec2𝜃 – tan2𝜃 =

(A)−1 (B) 0

(C) 1 (D) 2

91. 4cos3A - 3cosA =

(A) cos2A (B) cos3A

(C) sin3A (D) sin2A

92. 1 – cos2A =

(A) sin2A (B) 2sin2A

(C) cos2A (D) 2cos2A

93. tan2A =

(A) (B)
27

(C) (D)

94. fdlh ∆ABC esa cos =

( ) ( )
(A) (B)

( )
(C) (D)

In any ∆ABC, cos =

( ) ( )
(A) (B)

( )
(C) (D)

95. ¾

(A) tan3A (B) cot3A

(C) tan2A (D) cot2A

96. tan𝜃.sin2𝜃 =

(A) 1 + cos2𝜃 (B) 1 – cos2𝜃

(C) 1+ sin2𝜃 (D) 1 – sin2𝜃

97. sin( +𝜃) dk eku gksxk

(A) cos𝜃 (B) -cos𝜃

(C) sin𝜃 (D) -sin𝜃

28

The value of sin( +𝜃) is

(A) cos𝜃 (B) -cos𝜃

(C) sin𝜃 (D) -sin𝜃

98. sec(2700 + 𝜃) =

(A) cos𝜃 (B) cosec𝜃

(C) -cosec𝜃 (D) sin𝜃

99. sin(-650) =

(A) cos250 (B) –cos250

(C) sin250 (D) sin650

100. 2cos2450 + sin300 + cos00 – tan450 dk eku gksxk

(A) 1 (B) &1

(C) 2 (D) 0

The value of 2cos2450 + sin300 + cos00 – tan450 will be

(A) 1 (B) &1

(C) 2 (D) 0

29

[k.M&c @ Section-B

y?kq mÙkjh; iz”u @ Short Answer Type Questions.

iz'u la[;k 1 ls 30 y?kq mÙkjh; iz”u gSaA fdUgha 15 iz”uksa ds mÙkj nsaA izR;sd ds fy, 2

vad fu/kkZfjr gSA 15x2=30

Question Nos 1 to 30 are short Answer Type. Answer any 15 questions.

Each question carries 2 marks. 15x2=30

1- fl) djsa fd %&

cos(A+450) + sin(A-450) = 0

Prove that :-

cos(A+450) + sin(A-450) = 0 2

2- tan(1050) dk eku fudkysaA

Find the value of tan(1050) 2

3- fl) djsa fd %&

= cosec𝜃 - cot𝜃

Prove that :-

= cosec𝜃 – cot𝜃 2

4- fl) djsa fd %&
∝
= cot2∝
∝

30

Prove that :-
∝
= cot2∝ 2
∝

5- fl) djsa fd %&

– =0

Prove that:-

– =0 2

6- fl) djsa fd = 𝑐𝑜𝑠𝑥– 𝑠𝑖𝑛𝑥

Prove that = 𝑐𝑜𝑠𝑥– 𝑠𝑖𝑛𝑥 2

7- ;fn cot𝜃 = gks] rks cos2𝜃 dk eku Kkr djsaA

If cot𝜃 = , then find the value of cos2𝜃. 2

8- fl) djsa fd fcUnw,sa (6, 9), (0, 1), (-6, -7) lajs[kh gSA

Prove that the points (6, 9), (0, 1), (-6, -7) are collinear. 2

9- fl) djsa fd %&

+ = 2sec𝜃

Prove that :-

+ = 2sec𝜃 2

31

10- ;fn sinx = - vkSj 𝜋 < x < gks] rks sin2x dk eku Kkr djsa

If sinx = - and 𝜋 < x < , then find the value of sin2x. 2

11- sin7x – sinx dk eku Kkr djsaA

Find the value of sin7x – sinx. 2

12- fdlh f=Hkqt ∆ABC esa ;fn a = 2, b = √6 rFkk c= √3-1 gks] rks lcls NksVs

dks.k dks Kkr djsaA

In a ∆ABC if a = 2, b = √6 and c= √3-1, then find the smallest

angle. 2

13- fl) djsa fd %&

+ = 4cos2𝑥

Prove that :-

+ = 4cos2𝑥 2

14- fl) djsa fd %&

Sin3x= 3sinx - 4sin3x

Prove that :-

Sin3x= 3sinx - 4sin3x 2

15- fl) djsa fd

32

( )
=
( )

Prove that :

( )
= 2
( )

16- fl) djsa fd %&

cos( - x).cos( -y) –sin( -x).sin( -y) = sin (x+y).

Prove that :-

cos( - x).cos( -y) –sin( -x).sin( -y) = sin (x+y). 2

17- fl) djsa fd %&

√
sin750 + sin150 =

Prove that : -

√
sin750 + sin150 = 2

18- fcUnqvksa ¼8] 6½ vkSj ¼4] 2½ dks feykus okys js[kk[k.M ds e/; fcUnq dk fu;ked

Kkr djsaA

Find the coordinates of the midpoint of the line segment joining the

points (8, 6) and (4, 2). 2

19- fcUnqvksa ¼&5] 9½ rFkk ¼4] 3½ dks feykusokyh js[kk[k.M dks 5 % 2 ds vuqikr esa

vr% foHkkftr djus okyh fcUnq dk fu;ked Kkr djsaA

33

Find the coordinates of the point which divides the line segment

joining the points (-5, 9) and (4, 3) internally in the ratio 5:2. 2

20- fcUnq ¼&1] 7½] fcUnqvksa A¼&3] 9½ rFkk B¼2] 4½ dks feykus okyh js[kk[k.M dks

fdl vuqikr esa ck¡Vrk gS \

In which ratio does the point (-1, 7) divide the line segment joining

the points A(-3, 9) and B(2, 4) ? 2

21- fdlh vk;r ds rhu Øekxr 'kh"kZ (6, -1), (10, 6) vkSj (3, -5) gS rks pkSFkk

'kh"kZ Kkr djsaA

The three vertices of a rectangle taken in order are (6, -1), (10, 6)

and (3, -5), then find the fourth vertex.

2

22- y dk eku Kkr djsa ftlds fy, fcUnq A(3, y) vkSj B(4, 1) ds chp dh nwjh

√10 gSA

Find the value of y for which the distance between the points

A(3, y) and B(4, 1) is √10. 2

23- fl) djsa fd %&
( )
= cotA.cotB -1
.

Prove that :-

( )
= cotA.cotB -1 2
.

34

24- fl) djsa fd %&

2sinA.cosB – sin(A-B) = sin (A + B)

Prove that :-

2sinA.cosB – sin(A-B) = sin (A + B) 2

25- fl) djsa fd %&
( ) ( )
+ - = 3.
( ) ( )

Prove that :-

( ) ( )
+ - = 3. 2
( ) ( )

26- fl) djsa fd %&

Sin{(2n + 1)𝜋 – 𝜃} = sin𝜃

Prove that :-

Sin{(2n + 1)𝜋 – 𝜃} = sin𝜃 2

27- ;fn A rFkk B U;wu dks.k gsk rFkk sinA = rFkk sinB = gks] rks sin(A + B)

dk eku Kkr djsaA

If A and B are acute angles and sinA = and sinB = then, find the

value of sin(A + B). 2

28- fl) djsa fd %&

35

Cos180 – sin180 = √2cos270

Prove that :-

2
Cos180 – sin180 = √2cos270

29- fl) djsa fd %&
( )
tanA – tanB =
.

Prove that :-

( )
tanA – tanB = 2
.

30- fl) djsa fd %&

Sin(600 – A) = (√3cosA – sinA)

Prove that :-

Sin(600 – A) = (√3cosA – sinA) 2

nh?kZ mÙkjh; iz”u @ Long Answer Type Questions.

iz'u la[;k 31 ls 38 nh?kZ mÙkjh; iz”u gSaA fdUgha 4 iz”uksa ds mÙkj nsaA izR;sd ds fy, 5

vad fu/kkZfjr gSA 4x5=20

Question Nos 31 to 38 are Long Answer Type. Answer any 4 questions.

Each question carries 5 marks. 4x5=20

31- fl) djsa fd %&

=

36

Prove that :-

= 5

32- ;fn sec𝜃 – tan𝜃 = 𝑥 gks] rks fl) djsa fd cos𝜃 =

If sec𝜃 – tan𝜃 = 𝑥 then prove that cos𝜃 = 5

33- tan2250.cot4050 + tan7650.cot6750 dk eku Kkr djsaA

Find the value of tan2250.cot4050 + tan7650.cot6750. 5

34- ;fn cosecA + secA = cosecB + secB gks] rks fl) djsa fd %&

tanA.tanB = cot

If cosecA + secA = cosecB + secB, then prove that :-

tanA.tanB = cot 5

35- ;fn A + B + C = 𝜋 gks] rks fl) djsa fd %&

sin2A + sin2B + sin2C = 4sinA. SinB.sinC

If A + B + C = 𝜋, then prove that :-

sin2A + sin2B + sin2C = 4sinA. SinB.sinC 5

36- fl) djsa fd Øe ls fcUnq,¡ (6, 8), (3, 7), (-2, -2) vkSj (1, -1) ,d lekUrj

prqHkqZt ds 'kh"kZ gSA

Prove that the points (6, 8), (3, 7), (-2, -2) and (1, -1) taken in order

are the vertices of an parallelogram. 5

37- ;fn fdlh ∆ABC esa (a+b+c)(b+c-a) = 3bc gks] rks A dk eku Kkr djsaA
37

If in any ∆ABC (a+b+c)(b+c-a) = 3bc, then find the value of A. 5

38- /kjrh ij ,d ehukj m/ok/kZj [kM+h gSA /kjrh ds ,d fcUnq ls tks ehukj ds ikn

fcUnq ls 15 eh nwj gS] ehukj ds f”k[kj dk mUu;u dks.k 600 gSA ehukj dh

Å¡pkbZ Kkr djsaA

A tower stands vertically on the ground. From a point on the

ground, which is 15m away from the foot of the tower, the angle of

elevation of the top of the tower is found 600. Find the height of the

tower. 5

38