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Byju's Answer
Standard XII
Mathematics
Pythagorean Identities
The value of ...
Question
The value of
sec
−
1
(
1
4
10
∑
k
=
0
sec
(
7
π
12
+
k
π
2
)
sec
(
7
π
12
+
(
k
+
1
)
π
2
)
)
in the interval
[
−
π
4
,
3
π
4
]
equals
A
0
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B
0.00
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C
0.0
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Solution
sec
−
1
⎛
⎜ ⎜
⎝
1
4
10
∑
k
=
0
1
cos
(
7
π
12
+
k
π
2
)
cos
(
7
π
12
+
(
k
+
1
)
π
2
)
⎞
⎟ ⎟
⎠
=
sec
−
1
⎛
⎜ ⎜
⎝
1
4
10
∑
k
=
0
sin
[
(
7
π
12
+
(
k
+
1
)
π
2
)
−
(
7
π
12
+
k
π
2
)
]
cos
(
7
π
12
+
k
π
2
)
cos
(
7
π
12
+
(
k
+
1
)
π
2
)
⎞
⎟ ⎟
⎠
(
∵
sin
(
A
−
B
)
=
sin
A
⋅
cos
B
−
cos
A
⋅
sin
B
)
=
sec
−
1
⎛
⎜ ⎜
⎝
1
4
10
∑
k
=
0
(
sin
(
7
π
12
+
(
k
+
1
)
π
2
)
cos
(
7
π
12
+
k
π
2
)
)
−
(
cos
(
7
π
12
+
(
k
+
1
)
π
2
)
sin
(
7
π
12
+
k
π
2
)
)
cos
(
7
π
12
+
k
π
2
)
cos
(
7
π
12
+
(
k
+
1
)
π
2
)
⎞
⎟ ⎟
⎠
=
sec
−
1
(
1
4
10
∑
k
=
0
(
tan
(
7
π
12
+
(
k
+
1
)
π
2
)
)
−
(
tan
(
7
π
12
+
k
π
2
)
)
)
=
sec
−
1
1
4
(
tan
(
7
π
12
+
π
2
)
−
tan
(
7
π
12
)
+
tan
(
7
π
12
+
2
π
2
)
−
tan
(
7
π
12
+
π
2
)
+
.
.
.
.
+
tan
(
7
π
12
+
11
π
2
)
−
tan
(
7
π
12
+
10
π
2
)
)
=
sec
−
1
1
4
(
tan
(
7
π
12
+
11
π
2
)
−
tan
7
π
12
)
=
sec
−
1
1
4
(
tan
(
6
π
+
π
12
)
−
tan
(
π
−
5
π
12
)
)
=
sec
−
1
1
4
(
tan
π
12
+
tan
5
π
12
)
=
sec
−
1
1
4
[
(
2
−
√
3
)
+
(
2
+
√
3
)
]
=
sec
−
1
(
1
)
=
0
Suggest Corrections
0
Similar questions
Q.
The value of
sec
−
1
(
1
4
10
∑
k
=
0
sec
(
7
π
12
+
k
π
2
)
sec
(
7
π
12
+
(
k
+
1
)
π
2
)
)
in the interval
[
−
π
4
,
3
π
4
]
equals
Q.
The value of
sec
−
1
(
1
4
10
∑
k
=
0
sec
(
7
π
12
+
k
π
2
)
sec
(
7
π
12
+
(
k
+
1
)
π
2
)
)
in the interval
[
−
π
4
,
3
π
4
]
equals
Q.
If
tan
−
1
[
20
∑
k
=
0
sec
(
5
π
12
+
k
π
2
)
sec
(
5
π
12
+
(
k
+
1
)
π
2
)
]
=
−
tan
−
1
a
, then the value of
a
is
Q.
The value of
∑
13
k
=
1
1
s
i
n
(
π
4
+
(
k
−
1
)
π
)
6
)
s
i
n
(
π
4
+
k
π
6
)
is equal to
Q.
The value of
∑
13
k
=
1
1
s
i
n
(
π
4
+
(
k
−
1
)
π
)
6
)
s
i
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(
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+
k
π
6
)
is equal to
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Pythagorean Identities
Standard XII Mathematics
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