TRIGONOMETRY (WORKSHEET-CW1)

1 INFOMATHS/MCA/MATHS/

1.Ans. (c)If is a rational no.

Then As,

Now, as is rational

is rational .

 tan  is rational,

Similarly

 sin, cos  are rational numbers as well.

2.Ans. (a)

As

 is in IIIrd quadrant  sin  is negative.

[As, sin  is -ve]

 |sin  + cos | = - (sin + cos )

as in 3rd quadrant both sin, cos are –ve

 sin + cos  is –ve

3.Ans. (c)

dividing Nr/Dr by cos 37

= tan(82) = cot 8

4.Ans. (d)

Dividing Mr/Dr by cos 11

= tan56

5.Ans. (a)

Max value of sin + cos is at

 max value

is at

6.Ans. (a)

On squaring both sides:-

1 + sin2 = 2cos2 = 2 – 2sin2

1 + sin2 = 2 – 2sin2

2sin2 = 1 – sin2

cos  - sin = sin

7.Ans. (d)

Expanding determinant along R3

sin = cos2 + sin2

sin = 1

8.Ans. (a)(1 + tanA) (1 – tanB)

= 1 + tanA – tanB – tanA tanB …* Eq.

= ?

Also,

tan(A – B) = 1

tanA – tanB = 1 + tanA.tanB

tanA – tanB – tanAtanB = 1.

* EQ. – gives (1 + tanA) (1 – tanB) = 2

9.Ans. (b)As 1radian = 57 (approximately)

And we know sin is strictly increasing curve in

 sin 1 radian > sin1

10.Ans. (d)

in AMP in BMQ

h1 : h2 = 3 : 1

11.Ans. (a)tan2 = tan( +  +  - )

Also, and

as

So,

12.Ans. (b)sin2x = 1 – sinx

 sinx = 1 – sin2x = cos2x

 cos4x + cos2x = sin2x + sinx = 1

13.Ans. (b)cot-121 + cot-113 + cot-1 (-8)

= 

14.Ans. (a)sin ( cos ) = cos ( sin )

Put n = 0

15.Ans. (d)

16.Ans. (c)

{As cos ( - ) = - cos}

Using cosA. Cos2A. cos22A ....

We have,

here n = no. of factors.

A = smallest angle.

17.Ans. (b)Remember. In ABC

{Since, cosA + cosB + cosC }

{as }

18.Ans. (b)As sinx + sin2x = 1

sinx = 1 – sin2x

= cos2x

 cos12x + 3cos10x + 3cos8x + cos6x

= cos6x [cos6x + 3cos4x + 3cos2x + 1]

{as (a + b)3 = a3 + b3 + 3ab (a + b)}

= cos6x [(cos2x + 1]3 sin3x [sinx + 1]3

 (sin2x + sinx)3 {as a3.b3 = (ab)3}

 13 {as sin2x + sinx = 1}

19.Ans. (d)In ABE

y = x tan

In BEF

2h + y = xtan

Now, height of cloud above the surface

This can also be read as on replacing

Also and

20.Ans. (d)In BCD

h = a tan

in ADC

x + h = a tan

x + a tan  = a tan 

 x = a (tan  - tan )

21.Ans. (b)As

 Given equation can be written as

22.Ans. (c)As

23.Ans. (d)

24.Ans. (a)As

But

take

10t = 4 + 4t2

2t2 – 5t + 2 = 0

(2t – 1) (t – 2) = 0

,

25.Ans. ()sin30cos45 + cos3045= sin(30 + 45)

= sin 75

Using sin(A + B) = sinAcosB + cosA sinB.

26.Ans. (a)As B = 45, C = 105, c

Then third angle of ABC will be A = 30 {Since, sum of all angles of ABC = 180}

Now, using sine Rule:-

as

27.Ans. (b)a cos2 + bsin2

= a

28.Ans. (c)(no choice satisfies)

Dividing both sides by

i.e.

, where n  Z.

29.Ans. (c)As

30.Ans. (c)As sinx, cosx, tanx are in GP

sin.x tanx = cos2x

 sin2x = cos3x

Also cot2x = secx cot6x - cot2x = sec3x – secx

=secx (sec2x – 1) = secx.tan2x

= 1 {as sin2x = cos3x}

31.Ans. (d)As, the greatest angle of a  is opposite to the longest side always.

 Here x2 + x + 1 is the longest

Side (By observation)

 Angle opposite to this side will be largest.

{After solving the equation}

32.Ans. (d)2sin2 - 3sin - 2 = 0

2sin2 - 4sin + sin - 2 = 0

2sin(sin - 2) + (sin - 2) = 0

(2sin + 1) (sin - 2) = 0

, sin = 2

Whereas, sin = 2 is rejected.

33.Ans. (c)cosec A . [sinBcosC + cosB sinC]

= 1 As sin ( - ) = sin

34.Ans. (d)tan9 - tan27 - tan63 + tan81

= (tan9 + tan81) – (tan27 + tan63)

 4

35.Ans. (a)As

{ is in quad III

{as

36.Ans. (b)In ABC,

x = 60 cot 15

37.Ans. (c)sinx + cosx = 1

Dividing both sides by :-

x = 2n,

(check by taking a = even / odd in (c)

38.Ans. (c)sin = sin = n + (-1)n. 

= n + (-1)n

39.Ans. (a)As we know

cos  > sin  for 

cos  < sin  for 

and cos  = sin  for =

here for  = 10

cos  > sin 

 cos10 - sin10 is positive.

40.Ans. (c)8R2 = a2 + b2 + c2…* Equation

We know = R(circum radius)

a = 2R sin A

b = 2R sin B

c = 2R sin C

* Eq gives: - 8R2 = (2R sinA)2 + (2R sin B)2 + (2 Rsin C)2

R2.8 = 4R2 (sin2A + sin2B + sin2C)

2 = sin2A + sin2B + sin2C

 2 = 2(1 + cos A cos B cos C)

 cos A. 2cosB. cos C = 0

 either cosA = 0, or cos B = 0, cos C = 0

A = 90, B = 90, C = 90

41.Ans. (d)We know

2tan-1a – 2tan-1b = 2tan-1x

tan-1a – tan-1b = tan-1x

42.Ans. (a)

= 2 cos 60= 1

43.Ans. (d)...*Eq.

Given cot B – cot A = y … Eq. (1)

also, tanA – tan B = x

y = x. cot A . cot B

… Eq. (2)

From Eq. (1) and (2)

44.Ans. (a)cos ( - ) = a

cos ( - ) = b

Then sin2( - ) + 2ab cos ( - )

= sin2[( - ) - ( - )] + 2ab. cos [( - ) - ( - )]

= [sin ( - ) – cos ( - ) – cos ( - ) sin ( - )]2 + 2ab[cos (-) - cos ( - )+ sin ( - ) . sin ( - )]

= a2 (1 – b2) + b2(1 – a2) + 2a2b2= a2 + b2

45.Ans. (d)cos ( - ) = 1

 -  = 2n

cos ( + ) =

 +  = 2m cos-1 ()

2 = 2(m+n) cos-1 ()

 = (m+n)cos-1 ()

2 = 2(m-n) cos-1 ()

 = (m-n)cos-1 ()

(1)If m = 0, n = 0

 = -cos-1 ()= cos-1 ()

 two solutions as  = 

(2)If m = 1, n = 0

 = -cos-1 ()= - cos-1 ()

Also  =  ,  + ) = 2 cos-1

(3)If x = - 1, n = 0

, Also  -  = 0

 +  = - 2

 one solution

 four solutions are these

46.Ans. (b)

47.Ans. (a)tan( cos ) = cot ( sin )

Dividing both sides by

48.Ans. (a)

cot-1x + sin-1 =

cot-1x + tan-1=

cot-1x + tan-1+ tan-1(1) =

cot-1x + tan-1(3) =

So, x= 3

49.Ans. (a)

50.Ans. (c)a = 2, b = 3,

Using Since Rule:-

 sin B = 1.

B = 90 .

51.Ans. (a)= 0

52.Ans. (d)In ABC

x = 100 cot 60

Also, in  ACD

x + y = 100 cot 45

x + y = 100

But

 y = 100 – x

53.Ans. (c)a = 5, b , c = 5.

We know, the angle opposite to the longest side is largest. So B is the largest angle.

 B = 120

54.Ans. (b)

55.Ans. (d)Let the height of the tower be ‘h’ mts.

Then in ABC

m.

56.Ans. (a)Smallest angle is opposite to the smallest side.

 A = 20

57.Ans. (c)

58.Ans. (c)

Sol.x (x + 1)  0 and

x2 + x + 1  1

x2 + x  0

x (x + 1)  0

combing both equations.

x(x + 1) = 0

So the equation has two roots. i.e x = 0 and x = - 1.

59.Ans. (a)

5x = 1 – 6x2

6x2 + 5x – 1 = 0

6x2 + 6x – x – 1 = 0

6x(x + 1) – (x + 1) = 0

(6x – 1) (x + 1) = 0

60.Ans. (d)A = cos2 + sin4

= 1 – sin2 + sin4 = 1 – sin2 (1 – sin2)

= 1 – sin2.cos2

Also, - 1  sin2 1

0  (sin2)2 1

Adding 1 throughout :-

61.Ans. (d)By choices (a) choice fails because imaging roots

(b) gives  x = 0, as is not differ so fails

(c) gives x = -1, 1/2 again cos-1 (1 – (-1) = cos-12 not possible so (d) is as (1 – x) = 0

62.Ans. (b)sin4x+cos4x+sin2x+ = 0

sin4x+(1- sin2x)2+ sin2x+ = 0

sin4x + sin4x + 1- 2sin2x+ sin2x+ = 0

2sin4x- sin2x+1+ = 0

sin4x- + 1 + = 0

sin4x- + +1+ = 0

(sin2x - )2 = -

0 (sin2x - )2 1

0 -1

- - 

 (1+ )  -

 -

 only (b) choices satisfies

63.Ans. (d)acos2x + bsin2x + c = 0

acos2x + bsin2x = -C

= -C

Sin(2x + ) = -, where tan =

-1 <sin(2x + ) < 1

-1 < < 1

 can be satisfies by infinite triples (a, b, c)

64.Ans. (d)

Expanding the determinant along R.

= cosC . (-sinB tanA) – tanA . (sin B. cos C)

= - 2 tan A sin B cos C

65.Ans. (c)

There are four solutions.

66.Ans. (b)The formula 2sin'x holds for.

i) – 1  x  1

ii)

 2 sin-1x 

 sin-1x 

So,  x 

67.Ans. (d) ONE as

68.Ans. (c)cos-1x > sin-1x  sin-1x – cos-1x < 0

Also cos-1x > sin-1x for -1  x < 0

as

(2)

69.Ans. (a)

n = even  2x1 – 2x2 = n

 x1 – x2as n = even

 x1 – x2 = n, n  Z

70.Ans. (a)If A and B are position of person at lower and higher floor respectively and S is the position of statue, then by the above figure.

 AO = OS = 100…(i)

and

…(ii)

The distance between the two persons, AB = OB – OA

71.Ans. (a)As cot, cot2…….cotn = 1

cos cos2…….cosn =sin sin2……sinn

cos2 cos22…..cos2n=[sin21]

cos cos2…….cosn

72.Ans. (d)cos + cos  = a

sin + sin  = b....(*)

Also,

But  +  = 2( is A.M. of , )

Hence

73.Ans. (c)(1 + tan1)(1 + tan2)(1 + tan3) …..(1 + tan44) (1 + tan45) = 2n….. * Eq.

Considering

(1 + tan1) . (1 + tan44)

=1 + tan1 + tan 44 + tan1 tan 44 =1+(tan45)=2

{Since,

1 – tan 44 tan 1 = tan44+ tan1

tan44 + tan1 + tan44 tan 1 = 1. }

Now Eq. * gives 222 . (tan45 + 1) = 2n

223 = 2n

 n = 23.

74.Ans. (c)sin12 sin 4854

75.Ans. (a)

as sin (2 - ) = - sin 

So

So they will cancel out.

76.Ans. (b)a = 5, b = 4,

9 – 9 cosC = 7 + 7cos C

2 = 16cos C

5 = 41 – c2

c2 = 36

c = 6

77.Ans. (a)

Also, the triangle is right angled at A

 b2 + c2 = a2

= tan-1(1) 45

78.Ans. (d)3a = b+c

tantan=

= = as

= as b + c = 3a

79.Ans. (c)cos-1(cosx) = x

is true for all x [0, ]

Since, cos x  [-1, 1]

80.Ans. (c)

Since

81.Ans. (b)Let the height of tower be h.

 h = 100 mt.

82.Ans. (d)in  ABE

in  ACD,

h = 120 mt.

height of tower = 360 – 120 = 360 – 120 = 240 mt.

83.Ans. (c)Consider, a = 7, b = 5, c = 3.

 A = 120

84.Ans. (d)

X = 30

85.Ans. (b)We know, l = r

radians.

86.Ans. (b)radius of metallic ring = 1 foot [circumference = 2r = 2 foot] since the ring has been moulded into a circular arc ℓ = 2, radius = 80 ft.

radius

Area formed by the sector

= 80  . sq. feet.

87.Ans. (b)A, B, C, D are the angles of the cyclic quadrilateral

 A + C = B + D = 180

cosA + cosB + cosC + cosD

= (cosA + cosC) + (cosB + cosD)

= 0

88.Ans. (c)x.cos - ysin = 

x.sin + ycos = 

Squaring & adding:-

x2cos2 + y2sin2 - 2xysincos = 2

x2sin2 + y2cos2+ 2xysincos = 2

x2(cos2 + sin2) + y2(sin2 + cos2) = 2+ 2

x2 + y2 = 2+ 2

89.Ans. (b)Matrix A is non-invertible

If |A| = 0

i.e.

sin2 - cos2= 0

-cos2 = 0

cos2 = 0

90.Ans. (c)

91.Ans. (a)A + C = B

tan (A + C) = tan B

tanA + tanC = tanB . (1 – tanA.tanC)

tanA + tanC = tanB – tanA. tanB. tanC

 tanA . tanB. tanC = tanB – tanA – tanC

92.Ans. (c)Consider

x

But

93.Ans. (a)

As,  lies in 1st quadrant.

94.Ans. (b)sin2 - sin2 - 15cos2 = 0

1 – cos2 - sin2 - 15cos2 = 0

1 – sin2 - 16cos2 = 0

(cos - sin)2 – 16cos2 = 0

(cos - sin)2 = 16cos2

cos - sin =  4 cos.

cos  - sin = 4cos, cos - sin = - 4cos

- 3cos = sin5 cos = sin

tan = - 3 tan  = 5

 = m - tan-1 (3)  = n + tan-1(5)

95.Ans. (b)In a , if length of base is equal to the length of perpendicular.

Then the angle of evaluation is 45 always.

96.Ans. (b)tanA + cotA= 4

square both sides

tan2A+cot2A+2tanAcotA = 16

tan2A+cot2A= 16-2=14

Again squaring both sides

tan4A+cot4A+2tan2Acot2A =196

tan4A+cot4A = 196 – 2 = 194

97.Ans. (a)Let the sides be

a = x, b = 2x

 A = , B =  + 60

According to sin Rule:-

(say)

From the 1st and 2nd members:-

sin ( + 60) = 2sin

sin . cos60 + cos.sin60 = 2sin

 = 30, so other angle  + 60 = 90

 so triangle is right angle

98.Ans. (c)sinA = sinB, cosA = cosB

sin A = sinB

A = n + (-1)nB…………(i)

CosA = CosB

A= 2 n B……..(ii)

From (i) and (ii)

A = 2 n + B

99.Ans. (a)tan-1x + tan-1y

x + y = 1 – xy

x + y + xy = 1

100.Ans. (d)The equation 3cosx + 4sin x = 6 has no solution as the maximum value of 3cosx + 4sinx is 5.

i.e

Which is never equal to RHS.

 No solution does exist.

101.Ans. (a)sinx(1 + cosx) = z

= - sin2x + cosx + cos2x

= - (1 – cos2x) + cosx + cos2x

= 2 cos2x + cosx – 1

For max./min Put

2cos2x + cosx – 1 = 0

2cos2x + 2cosx – cosx – 1 = 0

2cosx (cosx + 1) – (cosx + 1) = 0

(2cosx – 1) (cosx + 1) = 0

2 cosx – 1 = 0, cosx = - 1

, cosx = - 1

x = 60,x = ,

Double derivative test:-

= - 4sinx.cosx – sinx.

gives maxima.

function f(x) is maximum at

102.Ans. (c)

103.Ans. (b)cos40 + cos80 + cos160

= (cos40 + cos80) + cos160

= 2cos60. cos20 + cos160

= cos20 + cos160

= cos20- cos20 = 0

as cos 160 = cos (180 - 20)

= - cos20

104.Ans. (d)2B = A + C, A+B+C = 180

3B = 180 B = 60

Also A + B + C = 

A =  - (B + C)

sin A = sin (B + C)

105.Ans. (d) Let the altitude of the aeroplane be _____.

 in AMCin  MCB,

h = x.

mt.

106.Ans. (c)

107.Ans. (c)cosec x + cotx = 2sinx,

1 + cosx = 2sin2x

2(1 – cos2x) = 1 + cosx

2 – 2cos2x = 1 + cosx

2cos2x + cosx – 1 = 0

(2cosx – 1) . (cosx + 1) = 0

- 1

, 

108.Ans. (d)ABCD is a cyclic quad.

A + C = 180 = 

sin(A + C) = sin = 0.

109.Ans. (a)max. value of

3cosx + 4sinx + 5 is

= 5 + 5  10.

110.Ans. (c)The greatest angle is always opposite to the longest side of a triangle.Among a, b and . The largest side is

 Using the cosine rule.

gives C = 120.

111.Ans. (a)

Take tan-1y = 

y = tan

Take

112.Ans. (b)= (let)

ac + a2 + b2 + bc = 2ab

= bc + ab + c2 + ac

a2 + b2 = c2 + ab

1 INFOMATHS/MCA/MATHS/

1 INFOMATHS/MCA/MATHS/