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CBSE Questions for Class 10 Maths Introduction To Trigonometry Quiz 10 - MCQExams.com
CBSE
Class 10 Maths
Introduction To Trigonometry
Quiz 10
csc
(
90
−
θ
)
=
is equivalent to
Report Question
0%
sec
θ
0%
sin
θ
0%
cos
θ
0%
None of these
Explanation
In the figure,
c
o
s
e
c
(
90
−
θ
)
=
r
x
and
sec
θ
=
r
x
So,
c
o
s
e
c
(
90
−
θ
)
=
sec
θ
Hence,
sec
θ
is correct.
Solve:
sec
70
∘
sin
20
∘
+
cos
20
∘
cosec
70
∘
Report Question
0%
0
0%
1
0%
−
1
0%
2
Explanation
Given:
sec
70
∘
sin
20
∘
+
cos
20
∘
cosec
70
∘
=
sec
70
∘
sin
(
90
∘
−
70
∘
)
+
cos
20
∘
cosec
(
90
∘
−
20
∘
)
=
sec
70
∘
cos
70
∘
+
cos
20
∘
sec
20
∘
[
∵
sin
(
90
∘
−
θ
)
=
cos
θ
and
cosec
(
90
∘
−
θ
)
=
sec
θ
]
=
1
+
1
[
∵
sec
θ
.
cos
θ
=
1
]
=
2
So,
2
is correct.
If
A
and
B
are complementary angles, then
sin
A
×
sec
B
=
Report Question
0%
1
0%
0
0%
−
1
0%
2
Explanation
Given that both the angles are complementary i.e.,
A
+
B
=
90
∘
So,
A
=
90
∘
−
B
We have :
=
sin
A
×
sec
B
=
sin
(
90
o
−
B
)
×
sec
B
=
cos
B
×
sec
B
[using
cos
θ
=
sin
(
90
o
−
θ
)
]
=
cos
B
×
1
cos
B
=
1
tan
5
∘
tan
25
∘
tan
30
∘
tan
65
∘
tan
85
∘
=
Report Question
0%
1
0%
1
2
0%
√
3
0%
1
√
3
Explanation
(
tan
5
∘
tan
85
∘
)
(
tan
25
∘
tan
65
∘
)
tan
30
∘
=
[
tan
5
∘
tan
(
90
∘
−
5
∘
)
]
[
tan
25
∘
(
tan
90
∘
−
25
∘
)
]
×
tan
30
∘
=
tan
5
∘
cot
5
∘
tan
25
∘
cot
25
∘
×
1
√
3
=
1
×
1
×
1
√
3
=
1
√
3
.
cot
(
90
∘
−
θ
)
is equivalent to
Report Question
0%
cot
θ
0%
cos
θ
0%
tan
θ
0%
−
tan
θ
Explanation
we know
cos
(
90
−
θ
)
=
tan
θ
So,
tan
θ
is correct.
cosec
(
75
∘
+
θ
)
−
sec
(
15
∘
−
θ
)
=
Report Question
0%
2
sec
θ
0%
2
c
o
s
e
c
θ
0%
0
0%
1
Explanation
Given,
csc
(
75
o
+
θ
)
−
sec
(
15
o
−
θ
)
=
csc
(
75
o
+
θ
)
−
sec
[
90
o
−
(
75
o
+
θ
)
]
=
csc
(
75
o
+
θ
)
−
csc
(
75
o
+
θ
)
=
0
0
will be the answer
If
tan
2
A
=
cot
(
A
−
21
∘
)
, where
2
A
is an acute angle, then
∠
A
=
____
Report Question
0%
24
∘
0%
27
∘
0%
35
∘
0%
37
∘
Explanation
Given,
tan
2
A
=
cot
(
A
−
21
∘
)
Therefore,
cot
(
90
∘
−
2
A
)
=
cot
(
A
−
21
∘
)
⇒
90
∘
−
2
A
=
A
−
21
∘
⇒
90
∘
+
21
∘
=
2
A
+
A
⇒
111
∘
=
3
A
⇒
A
=
37
∘
So,
∠
A
=
37
∘
is correct.
sin
2
20
∘
+
sin
2
70
∘
=
?
Report Question
0%
0
0%
1
0%
2
0%
3
Explanation
sin
2
20
∘
=
cos
2
(
90
∘
−
20
∘
)
=
cos
2
70
∘
Substituting
sin
2
20
∘
=
cos
2
70
∘
sin
2
20
o
+
sin
2
70
o
=
cos
2
70
∘
+
sin
2
70
∘
=
1
[
∵
sin
2
θ
+
cos
2
θ
=
1
]
Evaluate:
tan
35
∘
cot
55
∘
+
cot
78
∘
tan
12
∘
=
Report Question
0%
0
0%
1
0%
2
0%
None of these
Explanation
Given expression is
tan
35
o
cot
55
o
+
cot
78
o
tan
12
o
We can rewrite the expression as:
tan
35
o
cot
(
90
o
−
35
o
)
+
cot
78
o
tan
(
90
o
−
78
)
We know that,
cot
(
90
o
−
θ
)
=
tan
θ
and
tan
(
90
o
−
θ
)
=
cot
θ
∴
The expression becomes,
tan
35
o
tan
35
o
+
cot
78
o
cot
78
o
⇒
1
+
1
⇒
2
Hence,
tan
35
o
cot
55
o
+
cot
78
o
tan
12
o
=
2
If
sin
15
∘
=
cos
(
n
×
15
∘
)
, then
n
=
.
.
.
.
.
Report Question
0%
1
0%
2
0%
5
0%
0
Explanation
As we know that
cos
(
90
∘
−
θ
)
=
sin
θ
sin
15
∘
=
cos
(
90
∘
−
15
∘
)
=
cos
75
∘
=
cos
(
5
×
15
∘
)
So
n
=
5
cot
1
∘
cot
2
∘
.
.
.
.
.
.
cot
89
∘
=
Report Question
0%
1
2
0%
1
0%
0
0%
−
1
Explanation
cot
1
∘
cot
2
∘
.
.
.
.
.
.
cot
89
∘
=
(
cot
1
o
⋅
cot
89
o
)
(
cot
2
o
⋅
cot
88
o
)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
cot
44
o
⋅
cot
46
o
)
(
cot
45
o
)
=
(
cot
1
o
⋅
tan
1
o
)
(
cot
2
o
⋅
tan
2
o
)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
cot
44
o
⋅
tan
44
o
)
(
cot
45
o
)
, since
cot
(
90
o
−
θ
)
=
tan
θ
=
(
1
)
(
1
)
.
.
.
.
.
.
.
(
1
)
(
1
)
=
1
, since
tan
θ
⋅
cot
θ
=
1
Without trigonometric table, evaluate
cot
63
∘
tan
27
∘
Report Question
0%
0
0%
1
0%
2
0%
−
2
Explanation
Taking
cot
63
∘
tan
27
∘
cot
(
90
−
27
)
∘
tan
27
∘
We know,
cot
(
90
0
−
θ
)
=
tan
θ
tan
27
∘
tan
27
∘
=
1
Find the value of
tan
10
∘
tan
15
∘
tan
75
∘
tan
80
∘
Report Question
0%
1
16
0%
0
0%
1
0%
None of these
tan
1
∘
tan
2
∘
tan
3
∘
.
.
.
tan
89
∘
=
Report Question
0%
1
0%
−
1
0%
0
0%
None of above
Explanation
(
tan
1
∘
tan
89
∘
)
(
tan
2
∘
tan
88
∘
)
.
.
.
.
.
(
tan
44
∘
tan
46
∘
)
tan
45
∘
, group pair of terms
=
{
tan
1
∘
tan
(
90
∘
−
1
∘
)
}
{
tan
2
∘
tan
(
90
∘
−
2
∘
)
}
.
.
.
.
.
.
{
tan
44
∘
tan
(
90
∘
−
44
∘
)
}
tan
45
∘
=
(
tan
1
∘
cot
1
∘
)
(
tan
2
∘
cot
2
∘
)
.
.
.
.
.
(
tan
44
∘
cot
44
∘
)
tan
45
∘
=
(
1
×
1
.
.
.
.
.
1
)
×
1
=
1
Note that
tan
θ
×
cot
θ
=
1
Without trigonometric table, evaluate
sec
41
o
cosec
49
o
Report Question
0%
0
0%
1
0%
2
0%
−
2
Explanation
Given
sec
41
o
csc
49
o
=
sec
(
90
o
−
49
o
)
csc
49
o
=
csc
49
o
csc
49
o
as we know
sec
(
90
o
−
θ
)
=
csc
θ
=
1
Hence
1
is the answer
Find the value of
cos
2
θ
(
1
+
tan
2
θ
)
+
sin
2
θ
(
1
+
cot
2
θ
)
.
Report Question
0%
1
0%
3
0%
2
0%
4
Explanation
we know that ,
1
+
tan
2
θ
=
sec
2
θ
1
+
cot
2
θ
=
cosec
2
θ
Now,
cos
2
θ
(
1
+
tan
2
θ
)
+
sin
2
θ
(
1
+
cot
2
θ
)
=
cos
2
θ
sec
2
θ
+
sin
2
θ
cosec
2
θ
=
1
+
1
=
2
Therefore, Answer is
2
sin
81
∘
+
tan
81
∘
, when expressed in terms of angles between
0
∘
and
45
∘
, becomes
Report Question
0%
sin
9
∘
+
cos
9
∘
0%
cos
9
∘
+
tan
9
∘
0%
sin
9
∘
+
tan
9
∘
0%
cos
9
∘
+
cot
9
∘
Explanation
We know that,
sin
A
=
cos
(
90
˚
−
A
)
tan
A
=
cot
(
90
˚
−
A
)
sec
A
=
cosec
(
90
˚
−
A
)
Using these, we get
sin
81
∘
=
sin
(
90
∘
−
9
∘
)
=
cos
9
∘
tan
81
∘
=
tan
(
90
˚
−
9
)
=
cot
9
∘
∴
sin
80
∘
+
tan
81
∘
=
cos
9
∘
+
cot
9
∘
Find the value of :
cos
38
∘
csc
52
∘
tan
18
∘
tan
35
∘
tan
60
∘
tan
72
∘
tan
55
∘
=
Report Question
0%
√
3
0%
1
3
0%
1
√
3
0%
2
√
3
Explanation
Given:
cos
38
o
csc
52
o
tan
18
o
tan
35
o
tan
60
o
tan
72
o
tan
55
o
=
cos
38
o
csc
(
90
o
−
38
o
)
(
tan
18
o
×
tan
72
o
)
×
(
tan
35
o
×
tan
55
o
)
×
tan
60
o
=
cos
38
o
sec
38
o
tan
18
o
tan
(
90
o
−
18
o
)
tan
35
o
tan
(
90
o
−
35
o
)
tan
60
o
( we know that
tan
(
90
o
−
θ
)
=
cot
θ
and
cot
θ
=
1
tan
θ
and
sec
θ
=
1
cos
θ
and
tan
60
o
=
√
3
)
∴
cos
38
o
csc
52
o
tan
18
o
tan
35
o
tan
60
o
tan
72
o
tan
55
o
=
1
(
tan
18
o
cot
18
o
)
(
tan
35
o
cot
35
o
)
√
3
=
1
√
3
Evaluate:
tan
30
∘
cot
60
∘
Report Question
0%
1
√
2
0%
1
√
3
0%
√
3
0%
1
Explanation
Taking
tan
30
∘
cot
60
∘
tan
30
∘
cot
(
90
−
30
)
∘
as we know,
cot
(
90
0
−
θ
)
=
tan
θ
tan
30
∘
tan
30
∘
=
1
Find the name of the person who first produce a table for solving a triangle's length and angles.
Report Question
0%
William Rowan Hamilton
0%
Hipparchus
0%
Euclid
0%
Issac Newton
Explanation
Hipparchus
gave the first table of chords analogus to modern table of sine values, and used them to solve trigonometric problems
What is the meaning of trigonometry in Greek language?
Report Question
0%
Measurement
0%
Triangle Measure
0%
Angle Measure
0%
Degree Measure
Explanation
Trigonometry in the Greek language means triangle measure.
Trigono
⇒
triangle
+
metron
⇒
measure
So, option
B
is correct.
If
A
+
B
=
90
o
, then ......
Report Question
0%
sin
A
=
sin
B
0%
cos
A
=
cos
B
0%
tan
A
=
tan
B
0%
sec
A
=
cosec
B
Explanation
sec
A
=
r
x
and
cosec
A
=
r
x
So,
sec
A
=
cosec
A
cos
(
90
−
θ
)
sec
(
90
−
θ
)
tan
θ
cosec
(
90
−
θ
)
sin
(
90
−
θ
)
cot
(
90
−
θ
)
+
tan
(
90
−
θ
)
cot
θ
=
.
.
.
.
.
.
Report Question
0%
1
0%
−
1
0%
2
0%
−
2
Explanation
As we know that,
sin
(
90
o
−
θ
)
=
cos
θ
,
cos
(
90
o
−
θ
)
=
sin
θ
,
cosec
(
90
o
−
θ
)
=
sec
θ
,
cot
(
90
o
−
θ
)
=
tan
θ
,
sec
(
90
o
−
θ
)
=
cosec
θ
,
tan
(
90
o
−
θ
)
=
cot
θ
∴
cos
(
90
−
θ
)
sec
(
90
−
θ
)
tan
θ
cosec
(
90
−
θ
)
sin
(
90
−
θ
)
cot
(
90
−
θ
)
+
tan
(
90
−
θ
)
cot
θ
∴
sin
θ
×
1
cos
(
90
−
θ
)
×
tan
θ
1
sin
(
90
−
θ
)
×
sin
(
90
−
θ
)
×
tan
θ
+
cot
θ
cot
θ
=
sin
θ
×
1
sin
θ
1
cos
θ
×
cos
θ
+
1
=
1
+
1
=
2
Hence, option C is correct.
In the 5th century who created the table of chords with increasing 1 degree?
Report Question
0%
Hipparchus
0%
William Rowan Hamilton
0%
Euclid
0%
Ptolemy
Explanation
Ptolemy used length of chords to define his trigonometric functions
The value of
cot
1
∘
cot
2
∘
.
.
.
.
cot
89
∘
is .....
Report Question
0%
1
0%
0
0%
−
1
0%
None of above
Explanation
The given equation can be written as
(
cot
1
∘
cot
89
∘
)
(
cot
2
∘
cot
88
∘
)
.
.
.
.
.
(
cot
44
∘
cot
46
∘
)
cot
45
∘
=
(
cot
1
∘
cot
(
90
∘
−
1
∘
)
)
(
cot
2
∘
cot
(
90
∘
−
2
∘
)
)
.
.
.
.
.
(
cot
44
∘
cot
(
90
∘
−
44
∘
)
)
cot
45
∘
=
cot
1
∘
tan
1
∘
⋅
cot
2
∘
tan
2
∘
.
.
.
.
cot
44
∘
tan
44
∘
×
cot
45
∘
(Since
cot
(
90
∘
−
x
)
=
tan
x
)
)
=
1
×
1
.
.
.
.
.
1
×
1
(Since
cot
x
×
tan
x
=
1
,
cot
45
∘
)
=
1
Hence, option
A
is correct.
If
cos
θ
=
14
4
and
sin
θ
=
8
3
, what is the value of
cot
θ
?
Report Question
0%
11
16
0%
13
16
0%
16
21
0%
21
16
Explanation
cot
θ
=
cos
θ
sin
θ
=
14
4
8
3
=
14
4
×
3
4
cot
θ
=
21
16
So, option
D
is correct.
If
cos
θ
=
2
21
and
sin
θ
=
6
7
, what is the value of
tan
θ
?
Report Question
0%
4
0%
9
0%
6
0%
7
Explanation
tan
θ
=
sin
θ
cos
θ
=
6
7
2
21
=
6
7
×
21
2
tan
θ
=
9
So, option
B
is correct.
If
t
=
45
∘
, what is
sec
(
t
)
sin
(
t
)
−
c
o
s
e
c
(
t
)
cos
(
t
)
?
Report Question
0%
−
2
0%
−
1
0%
0
0%
π
2
0%
None of the above
Explanation
For
t
=
45
0
, the value of
sec
(
t
)
sin
(
t
)
−
csc
(
t
)
cos
(
t
)
may be determined as follows:
sec
(
t
)
sin
(
t
)
−
csc
(
t
)
cos
(
t
)
=
sec
(
45
0
)
sin
(
45
0
)
−
csc
(
45
0
)
cos
(
45
0
)
=
√
2
×
1
√
2
−
√
2
×
1
√
2
=
1
−
1
=
0
If
sin
θ
+
cos
θ
=
√
2
, find the value of
sin
θ
×
cos
θ
.
Report Question
0%
2
0%
√
2
−
1
0%
1
2
0%
2
sec
θ
Explanation
sin
θ
+
cos
θ
=
√
2
Squaring both sides, we get
⇒
(
sin
θ
+
cos
θ
)
2
=
2
⇒
sin
2
θ
+
cos
2
θ
+
2
sin
θ
.
cos
θ
=
2
⇒
1
+
2
sin
θ
.
cos
θ
=
2
⇒
2
sin
θ
cos
θ
=
1
⇒
sin
θ
cos
θ
=
1
2
.
Hence, the answer is
1
2
.
In the following figure
A
B
⊥
B
C
and
∠
A
C
B
=
30
∘
, given
B
C
=
√
300
m
. The length of
A
B
is
Report Question
0%
10
m
0%
100
m
0%
10
√
3
m
0%
100
√
3
m
Explanation
Given that:
From the fig;
A
B
⊥
B
C
,
∠
A
C
B
=
30
∘
,
B
C
=
√
3
00
m
To find:
A
B
=
?
Solution:
In
△
A
B
C
,
tan
∠
A
C
B
=
A
B
B
C
or,
tan
30
∘
=
A
B
B
C
or,
A
B
=
tan
30
∘
×
B
C
or,
A
B
=
1
√
3
×
√
300
m
or,
A
B
=
1
√
3
×
10
√
3
m
or,
A
B
=
10
m
Therefore, A is the correct option.
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