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INFOMATHS
OLD QUESTIONS-CW1
TRIGONOMETRY
1.
Let  be an angle such that 0 <  < /2 and tan (/2) is rational.
Then which of the following is true?
HCU-2012
(a) Both sin (/2) and cos(/2) are rational
(b) tan() is irrational
(c) Both sin () and cos() are rational
(d) none of the above
2.
If
5
3
 
, then
4
2
2cot  
14.
3.
1
sin 2 
15.
6.
(b) 1-cot 
(d) – 1+cot 
16.
17.
18.
2 cos, then cos– sinis equal to
PU CHD-2012
(c)

2 sin
sin 

(B)
19.
2 sec 
(D)
2
7.
If sin  =
cos 
 sin  0
sin 
cos 
0
0
0
1
then  =
(a)
(b) 


(c)

4
8.
If A – B
9.
(a) 2
(b) 1
(c) 0
Which of the following is correct?
4
(d)

2
20.
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21.
(d) 3
10.
11.
(c) sin 1 = sin 1
(d)
sin1 

180
4
56
33
(b)
(c)
16
63
5
, then tan A is :
2
1
(a) x 
8
1
1
1
(b) x 
(c) x 
(d) x 
8
2
2
If sinx + sin2x = 1, then the value of cos12 x + 3cos10x + 3cos8x +
cos6x is
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(a) – 1
(b) 1
(c) – 2
(d) 2
If the angle of elevation of a cloud at a height h above the level of
water in a lake is  and the angle of depression of its image in the
lake is , then the height of the cloud above the surface of the lake
is not correct:
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h  tan   tan  
(b)
tan   tan 
h  cot   cot  
(d)
cot   cot 
x 3
(b)
(c) x = 1
22.
13.
(a) 0
(b) 1
(c) 2/3
(d) – 1
The value of cot-1 (21) + cot-1 (13) + cot-1 (-8) is
sin     
h cos    
sin     
If
x
1
3
(d) x = 0
1 
sin 1 x  cot 1    ,
2 2
then x is :
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33
56
If sin2x = 1 – sinx, then cos4x + cos2x =
h sin    
If the angles of elevation of the top and bottom of a flag staff fixed
at the top of a tower at a point distant a from the foot of a tower
are  and , then height of the flag staff is :
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(a) a (sin  - sin )
(b) a (cos  - cos )
(c) a (cot  - cot )
(d) a (tan  - tan )
The solution of the equation
(b)
2
(c)
1
23.
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The value of
(a)

3
cos 1
(b)

4
(d)
5
3
12.
4
3
1
1
(c)
(d)
8
4
A
B
C
If A + B + C =  and x  sin sin sin , then :
2
2
2
(a) 0
(d)

1
(b) 
4
(a)
, then tan (2α) =
63
65
(d)
3
(b)
5
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(a)
1
4
BHU-2012
4
5
,
If cos     
and sin     
5
13


is :
sin1
If two towers of heights h1 and h2 subtend angles 60 and 30
respectively at the midpoint of the line joining their feet, then h1 :
h2 is
NIMCET-2012
(a) 1 : 2
(b) 1 : 3
(c) 2 : 1
(d) 3 : 1
0    
A  cot A 
(c)
 1  x2 
 2x 
 2x  
3sin 1 
 4cos 1 
 2 tan 1 

2 
2 
2 
 1 x 
 1 x  3
 1 x 
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(b) sin 1 < sin 1
1
3
1
(a) 
8
(c)
, then (1 + tan A) (1 – tan B) is equal to
(a) sin 1 > sin 1

20
15
(c)
(d)
16
21

4
5
The value(s) of cos cos
is (are):
cos
7
7
7
(a)
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
3
If cosec
(b)
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PU CHD-2012
(A) cos 34° (B) sin 34° (C) cot 56° (D) tan 56°
The maximum value of sin(x + /6) + cos(x + /6) in interval (0,
/2) is attained at
(A) /12
(B) /6
(C) /3
(D) 2
(A)
3
4
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sec 37
(c) cot 8
(d) tan 16
csc 37
cos11  sin11
The value of

cos11  sin11
If cos+ sin=

2
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(a) tan274 (b)
5.

4
(a)
9
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HCU-2012
4.
(d)
If sin (cos) = cos (sin), then sin 2 =
(a)
cos 37  sin 37
is
cos 37  sin 37
The value of
(c) 8
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is equal to
(a) 1+cot 
(c) – 1 – cot 
(b) π
(a) 0
3
2
2
6 1
 cos 1
is equal to :
3
2 3
(c)

2
(d)

6
BHU-2012
NIMCET-2012
1
INFOMATHS/MCA/MATHS/OLD QUESTIONS
INFOMATHS
24.
If
4
 x
, then the value of tan   is
5
2
sin  x  
(a)
36.
HCU-2011
25.
(a) ½ or 2
(b) ½ or – 2
(c) ¾ or – 2
(d) ¾ or 2
The value of sin 30° cos 45° + cos 30° sin 45°
[no correct answer was given in choices, correct answer should
3 1
be
]
2 2
(a)
26.
1 3
2
1 3
(b)
2 2
3
2
(d)
3
(a) A = 30°, a =
3 1 , b =
(b) A = 30°, a =
3 1 , b =
(c) B = 30°, a =
1 3 , b =
3 1 , b =

2
2
2
2
37.

3  1
3  1
3  1
3 1
38.
28.
(b) a
The general solution of
2n 
(b)
6
(c) No solutions
29.
The value of
(d)
1  tan 2 15
1  tan 2 15
2n 
n 
30.
31.
32.
3
(b)

(a)
n   1
3
2

(b)
42.
43.
(d) 2
34.
(b)
a
c
n   1
n
44.
45.

2
n 7
(d) n   1
6
(c) 1
46.
35.
If
x  n   1
(d)
x  n   1
n
(c) 3

4
n

4


4
, n  0, 1, 2,....
, n  0, 1, 2,....
If sin  = sin , then the angle  and  are related by
2
2a
2x
1 1  b

cos
 tan 1
2
2
1 a
1 b
1  x2
then x is equal to
ab
(c)
1  ab
(b) b
a b
(d)
1  ab
The value of 3 cot 200 -4 cos 200 is
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(a) 1
(b) -1
(c) 0
(d) N.O.T
If tan A – tan B = x and cot B – cot A = y, then cot (A – B) is
equal to
KIITEE-2010
1
y
x
(b)
1
xy
(c)
1 1

x y
(d)
1 1

x y
(e) None of these
If cos ( – ) = a, and cos ( – ) = b, then sin2 ( – ) +
2ab cos ( – ) is equal to
KIITEE-2010
(a) a2 + b2 (b) a2 – b2 (c) b2 – a2 (d) – a2 – b2
(e) None of these
The number of ordered pairs (,) where ,  (-,) satisfying
cos     
1
e
is
KIITEE-2010
(a) 0
(b) 1
(c) 2
(d) 4
(e) None of these
If tan  = (1 + 2-x)-1, tan  = (1 + 2x+1)-1, then  +  equals
KIITEE-2010
(a)

6
(b)

4
(c)

3
(d)

2
(e) None of these
(d) 0
47.
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(b) 2
m
, n = 0, 1, 2, …..
cos ( - ) = 1 and
The value of tan 9 - tan 27 - tan 63 + tan 81 is
(a) 1

2
(c)
(a)
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c
a
3 1
The general solution of the trigonometrical equation sinx + cosx =
1 is given by
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(a) x = 2n, n = 0, 1,  2, …
(a) a
(c)
3 1
(d)
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In a  ABC, cosec A(sin B cos C + cos B sinC) equals
(a)
3 1
m
is
6
n 5
(c) n   1
6
33.
3 1
 3 1

 60m
 3 1 
(b)
6
If sin x, cos x and tan x are in GP, then the value of cot 6x – cot2x
is:
NIMCET-2011
(a) 2
(b) – 1
(c) 1
(d) 0
The greatest angle of the triangle whose three sides are x2 + x + 1,
2x + 1 and x2 – 1 is
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(1) 150°
(2) 90°
(3) 135°
(4) 120°
The general value of θ satisfying the equation 2sin2 θ – 3sin θ – 2
= 0 is
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n
 3 1 

 60m
 3 1
41. If sin-1
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(a) 1
10
3
NIMCET-2011
(a)
10

a
(d)
ab
3 cos x  sin x  3 is:

7
(d)
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(a)  = 2n + (-1)n 
(b)  = n  
(c)  = n + (-1)n 
(d)  = (2n + 1)  + 
39. The value of cos 10 - sin 10 is
BHU-2011
(a) positive
(b) negative
(c) 0
(d) 1
40. In a triangle ABC, R is circumradius and
2
2
2
2
8R = a +b +c . The triangle ABC is
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(a) Acute angled
(b) Obtuse angled
(c) Right angled
(d) N.O.T
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(a) b
3
(c)
10
(b) x = 2n +
b
If tan   , then the value of a cos 2θ + b sin 2θ is
a
a
(c)
b
10
(c)
The solution of Δ ABC given that B = 45°, C = 105° and c = 2
is
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(d) B = 30°, a =
27.
(c)
2
1
(b)
From the top of a lighthouse 60 m high with its base at the sealevel, the angle of depression of a boat is 15. The distance of the
boat from the foot of the lighthouse is
BHU-2011
(a)
NIMCET-2011
1



cos    
4

equal to
(d) 4
1
3
3 
sin           , then the value of cos 
2
5
2 
If tan (cos) = cot(sin), then the value of
(a)
is
1
is
KIITEE-2010
(b)
1
2
2 2
(c)
1
3 2
(d)
1
4 2
(e) None of these
BHU-2011
2
INFOMATHS/MCA/MATHS/OLD QUESTIONS
INFOMATHS
48.
Solution of the equation
cot 1 x  sin 1
1

5

59.
is
4
KIITEE-2010
(a) x = 3
(b)
(c) x = 0
(e) None of these
49.
50.
x
1
60.
5
(d) None of these
5
2

cot  cos ec 1  tan 1 
KIITEE-2010
3
3

6
3
4
5
(a)
(b)
(c)
(d)
17
17
17
17
2
In ABC, a = 2, b = 3 and sin A 
then B is equal to
3
61.
62.
(a) 30
(b) 60
The value of
 
 
sin    cos    tan 3  45 
6
 
3
is
(c)
PGCET-2010
(a) 0
(b) 1
2
(c)
(d)
2
3
52.
63.
3
From a point 100 meters above the ground, the angles of
depression of two objects due south on the ground are 60 and
45. The distance between the object is
PGCET-2010
50
(a)

100
(c)
 mts
3 1
3

75
(b)

64.
3
 mts
53.
54.
55.
(b)
(c)
75
3
125
mts
(b)
mts
(d)
3
56.
tan A
0
sin B
0
tan A

2sin 1 x  sin 1 2 x 1  x2
3
100
If
(b) 90
is
(c) 75
mts
(b)
n 1
n2
n
n2
The value of
68.
(a) 17/6
(b) 7/16
If cos-1 x > sin-1x, then
mts
(c)
69.
(b) –1 < x < 0
1
0 x
tan 1 x( x  1)  sin 1 x 2  x  1 
(a) zero
(b) one
(c) two

2
(d)
1 x 
Consider the function
1
2


f  x   sin  2 x  
3

on R. Let x1 and
x2 be two real values such that f(x1) = f(x2). Then x1 – x2 is always
of the form
Hyderabad Central University – 2009
(a) n : n  Z
(b) 2n : n  Z
(c)
70.
n 1
n2
The number of solutions for
(KIITEE – 2009)
(d) None of these
(c) 6/17
2
(PGCET paper – 2009)
(d) 15
(d)
 1 1 
x
,

 2 2
(KIITEE – 2009)
(MCA : NIMCET – 2009)
(c)
holds for
(d)
(a) x < 0
1
1
 tan 1
 tan 1
1 2
1  (2)(3)
1
1
then  is equal to
 .....  tan 1
1  (3)( 4)
1  n(n  1)
n
n 1
(b)
67.
  tan 1
(a)

 3 3 
x
,

2 
 2

 2 

4

tansin 1    cos 1 

5



3




(c) x  [0, 1]
 1 3
50
(KIITEE – 2009)
has the value
sin B cos C
(a) x  (-1, 0)
3
b  6  2, c  2  3
58.
cos C
The smallest angle of a ABC whose sides are a = 1,
(a) 20
57.
The number of values of the triple t(a, b, c) for which a cos 2x + b
sin2x + c = 0 is satisfied by all real x is (MCA : KIITEE – 2009)
(a) 0
(b) 2
(c) 3
(d) infinite
(MCA : KIITEE – 2009)
(c) 1  3
(d)  1 3
The elevation of the tower 100 meters away is 30. The length of
the tower is
(PGCET paper – 2009)
(a)
(d) – 1    1
The formula
 
 
sin   cos   tan 3 45
3
 
6
1 3
(b) – 3   1
66.
(d) None of these
(PGCET– 2009)
(a)
1
1
 
2
2
3
1
  
2
2

65.
The greatest angle of ABC whose sides are a = 5, b  5 3 and
c = 5, is
PGCET-2010
(a) 45
(b) 100
(c) 120
(d) 60
The value of
(d)
13
 A 1
16
3
 A 1
4
(a) sin A sin B cos C
(b) 0
(c) 1
(d) None of these
The number of solution of |cos x| = sin x, 0  x  4, is
(MCA : KIITEE – 2009)
(a) 8
(b) 2
(c) 4
(d) None of these
3 1
3
, then x is
(b)
3
13
 A
4
16
0
 mts
3 1
4
If sin-1 x + cos-1 (1 – x) = sin-1 (-x), then x satisfies the equation
(MCA : NIMCET – 2009)
(a) 2x2 – x + 2 = 0
(b) 2x2 – 3x = 0
(c) 2x2 + x – 1 = 0
(d) None of these
The equation sin4x + cos4x + sin2 x +  = 0 is solvable for
(MCA : NIMCET – 2009)
(a)
(d) 120

(MCA : NIMCET – 2009)
(a) 1/6
(b) 1/3
(c) ½
(d) ¼
If A = cos2 + sin4, then for all values of 
(MCA : NIMCET – 2009)
(c)
KIITEE-2010
51.
tan 1 2 x  tan 1 3x 
(a) 1  A  2
The value of
(c) 90
If
(MCA : NIMCET – 2009)
(d) infinite
71.
3

3
:nZ
(d)
n 

3
:nZ
Two persons are standing at different floors of a tall building and
are looking at a statue that is 100 metres far from the building.
Angle of inclination of the person at higher floor is 60 and that of
the person at lower floor is 45. What is the distance between the
two persons?
Hyderabad Central University – 2009
(a)
is
2n 


3  1 100
(b)


3  1 100
(c) 3 100
(d) 100 / 3
The maximum value of (cos 1) (cos 2) …. (cos n) where 0 
1, 2, n  /2 and (cot 1) (cot 2) … (cot n) = 1 is
INFOMATHS/MCA/MATHS/OLD QUESTIONS
INFOMATHS
NIMCET – 2008
87.
If cos  + cos  = a, sin  + sin  = b and  is the arithmetic mean
between  and , then sin 2 + cos 2 is equal to
NIMCET – 2008
88.
(a)
72.
(a)
(c)
73.
74.
75.
1
2n / 2
(b)
1
1
2n
(c)
2n
(a  b) 2 /(a 2  b 2 )
(d) 1
(a  b) 2 /(a 2  b 2 )
(b)
a2  b2
(d) None
a2  b2
The value of
sin

n
 sin
3
5
 sin
 ...n
n
n
(a) 0
77.
(b) 6

4
terms is equal
(d) None
tan
C
7
then the side c is

91.
2
9
MCA : KIITEE – 2008
(d) None
(c) 2
In a  ABC, A = 90. Then
(a)
78.
n
2
(c)
In a  ABC, a = 5, b = 4 and
(a) 3
90.
MCA : KIITEE – 2008
(b) 1
tan 1
92.
b
c
 tan 1
ac
ab
(b)
If in a  ABC, 3a = b + c then
(c)
tan

2
B
C
, tan
2
2
(b)
tan
4
2
(c) 2
79.
cos-1 (cos x) = x is satisfied by
(a) x  R
(c) x  [0, ]
80.
The value of
2 tan 1
93.
(a) 
(b)
81.
82.
83.
84.
85.
86.

2
(c)

4
has no
(d) 75
 cos  
then sin  f() + cos 
sin  
ICET – 2007
1
0

0
(d) 
1
(b)
1 0 
0 1 


0
1 
1
0 
If A + C = B, then, tan A tan B tan C =
ICET – 2005
(a) tan B – tan A – tan C
(b) tan B + tan A – tan C
(c) tan B – tan A + tan C
(d) tan A + tan B + tan C
If a flag staff of 6 metres height, placed on the top of a tower
If
sin  
(a)
15
15 cot  17 sin 
, then, for 0 <  < 90
8 tan  16 sec
17
23
49
(b)
22
49
(c)
18
49
(d)
17
49
The general solution of the equation sin2 – sin2 - 15cos2 = 0 is
given by  equals
IP University : Paper – 2006
(a) n + tan-1 3 or m - tan-1 5
(b) n - tan-1 3 or m + tan-1 5
(c) n - tan-2 2 or m + tan-1 6
(d) n - tan-1 7 or m - tan-1 3
(e) None of these
95. When the length of the shadow of a pole is equal to a height of the
pole, then the elevation of source of light is
Karnataka PG-CET Paper – 2006
(a) 30
(b) 45
(c) 60
(d) 75
96. If tan A + cot A = 4 then tan4 A + cot4 A is equal to
Karnataka PG-CET Paper – 2006
(a) 110
(b) 194
(c) 88 (d) 194
97. If one side of a triangle is double of another side and the angle
opposite to these sides differ by 60, then the triangle is
Karnataka PG-CET Paper – 2006
(a) right angled
(b) an obtuse angled
(c) an acute angled
(d) None of these
98. If sin A = sin B and cos A = cos B, then
Karnataka PG-CET Paper – 2006
(a) A = n + B
(b) A = n - B
(c) A = 2 n + B
(d) A = 2n  - B
99. If tan-1 x + tan-1 y = /4, then
Karnataka PG-CET Paper – 2006
(a) x + y + xy = 1
(b) x + y – xy = 1
(c) x + y + xy + 1 = 1
(d) x + y – xy + 1 = 0
100. The equation 3 cos x + 4 sin x = 6 has _____ solution
Karnataka PG-CET Paper – 2006
(a) finite
(b) infinite
(c) one
(d) no
101. The value of sin x(1 + cos x) is maximum at:
MP: MCA Paper – 2004
(a) /3
(b) /2
(c) /6
(d) 3/4
94.
KIITEE – 2008
(d) None
A tower casts a shadow 100, long when the elevation of a source
of light is at 45. What is the height of the tower?
KARNATAKA – 2007
(a) 100 3 (b) 100m
(c) 10m
(d) 10 3 m
From the top of a light house 360 m height, the angles of
depression of the top and bottom of a tower are observed to be 30
and 60 respectively. What is the height of the tower?
KARNATAKA – 2007
(a) 200m
(b) 210m
(c) 190m
(d) 240m
The greatest angle of a triangle with sides 7, 5 and 3 is
KARNATAKA – 2007
(a) 60
(b) 90
(c) 120
(d) 135
For a triangle XYZ, if X
2
ICET – 2005
KIITEE – 2008
(b) x [-1, 1]
(d) None
is


sec  
1
ICET – 2007
(c) 60
is equal to
(d) None
1
1
 tan 1
3
7
(c)
(d) None
KIITEE – 2008
(a) 1
 sin 
f ( )  
cos 


f  
2

1

1

(a) 

 1 1 
It
sin 2 
A
 1
throws a shadow 2 3 of metres along the ground, then, the
angle in degrees that the sun makes with the ground is
ICET–2005
(a) 30
(b) 45
(c) 60
(d) 75
MCA : KIITEE – 2008
a
tan 1
bc
If (0 <  < 90 and the matrix
inverse than 
(a) 30
(b) 45
If (1 + tan 1) (1 + tan2) … (1 + tan 45) = 2n, then the value of n
is
NIMCET – 2008
(a) 21
(b) 22
(c) 23
(d) 24
The value of sin 12 and 48 sin 54
NIMCET – 2008
(a) sin 30 (b) sin230 (c) sin330 (d) cos3 30
to
76.
89.
If A, B, C, D are angles of a cyclic quadrilateral then cos A + cos
B + cos C + cos D is
KARNATAKA – 2007, UP-2002
(a) 1
(b) 0
(c) 2
(d) 3
If x cos  - y sin  =  and x sin  + y cos  then x2 + y2
ICET – 2007
(a) 2 (b) 2 (c) 2 + 2 (d) 2 – 2
 2 , Y = 2, Z  3  1 then  X is
KARNATAKA – 2007
(a) 45
(b) 60
(c) 75
(d) 30
A wire of length 20 cm is bent so as to form an arc of a circle of
radius 12 cm. The angle subtended at the center is
KARNATAKA – 2007
(a) 3/5 radians
(b) 5/3 radians
(c) 1/3 radians
(d) 5 radians
A circular metallic ring of radius 1 foot is reshaped into a circular
arc of radius 80 ft. The area of the sector formed is
KARNATAKA – 2007
(a) 20 sq ft.
(b) 40 sq. ft
(c) 80 sq. ft
(d) 60 sq. ft
102.

3  
tan  tan 1    
4 4

UPMCAT : Paper – 2002
(a) 117
(b) 3/7
(c) -1/7
(d) None of these
103. Cos40 + Cos80 + Cos 160 is equal to :
4
INFOMATHS/MCA/MATHS/OLD QUESTIONS
INFOMATHS
(a) -1
(b) 0
104. A, B, C are in A.P. b:c
(c) 1
UPMCAT : Paper – 2002
(d) N.O.T.
109. The maximum value of 3cosx + 4sinx + 5 is:
UPMCAT : Paper – 2002
(a) 10
(b) 0
(c) 5
(d) None of these
 3  1:1 then  A is equal to :
UPMCAT : Paper – 2002
(a) 103.5 (b) 98.5
(c) 101.5 (d) None of these
105. If two stones are 500 meters apart. The, angle of depressions
being 30 and 45 as seen by aeroplane what is the altitude the
plane is flying:
UPMCAT : Paper – 2002
110. The sides of a triangle are a, b and
greatest angle is :
(a) 60
(b) 90
(c) 120
111. sin[cot-1 cos(tan-1 y)] is equal to :
, then the
UPMCAT : Paper – 2002
(d) None of these
UPMCAT : Paper – 2002
(b) 250 3 mts
250 3  1 mts
(c) 250 3  1 mts
(d) None of these

 1

106. tan  tan  2 x    is equal to :
4

(a)
y 1
y2  2
(b)
(c)
y2
y 3
(d) None of these
(a)
2
UPMCAT : Paper – 2002
(a)
a2  ab  b2
2x 1
2x  1
2x 1
(b)
(c)
(d) None of these
2x  3
2x 1
2x  1
112. If
107. If cosec x + cot x = 2 sin x, where 0 ≤ x ≤ 2π In then:
UPMCAT : Paper – 2002
(a) x = π/3, 5 π/3
(b) x = π/3, 5π/6
(c) x = π/3, π
(d) None of these
108. In a cyclic quadrilateral ABCD, sin (A + C) is equal to :
UPMCAT : Paper – 2002
(a) ½ (b) 1
(c) – 1
(d) 0
y2 1
y2  2
1
1
3


then  C is:
bc ca abc
(a) 90
(b) 60
(c) 30
UPMCAT : Paper – 2002
(d) 45
ANSWERS (OLD QUESTIONS-CW1)
1
C
11
A
21
B
31
D
41
D
51
A
61
D
71
A
81
B
2
A
12
B
22
C
32
D
42
A
52
D
62
B
72
D
82
D
3
C
13
B
23
D
33
C
43
D
53
C
63
D
73
C
83
C
TRIGONOMETRY
4
5
6
7
D
A
A
D
14
15
16
17
A
D
C
B
24
25
26
27
A
C
A
B
34
35
36
37
D
A
B
C
44
45
46
47
A
D
B
A
54
55
56
57
B
D
A
C
64
65
66
67
D
C
B
D
74
75
76
77
C
A
B
A
84
85
86
87
D
B
C
B
8
A
18
B
28
C
38
C
48
A
58
C
68
C
78
D
88
C
9
B
19
D
29
C
39
A
49
A
59
A
69
A
79
B
89
B
91
A
101
A
111
A
10
D
20
D
30
C
40
C
50
C
60
D
70
A
80
C
90
C
5
92
C
102
C
112
B
93
A
103
B
94
B
104
C
95
B
105
D
96
B
106
C
97
A
107
C
98
C
108
D
99
A
109
A
100
D
110
C
INFOMATHS/MCA/MATHS/OLD QUESTIONS