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Standard XI
Mathematics and Statistics
Question
The value of
∑
88
n
=
0
1
c
o
s
n
k
.
c
o
s
(
n
+
1
)
k
, where k is
1
o
is equal to
s
i
n
2
k
c
o
s
k
c
o
s
k
c
o
s
e
c
2
k
none of these
c
o
s
2
k
s
i
n
k
A
s
i
n
2
k
c
o
s
k
B
c
o
s
k
c
o
s
e
c
2
k
C
c
o
s
2
k
s
i
n
k
D
none of these
Open in App
Solution
Verified by Toppr
Multiply & divide by sin k
T
n
=
1
s
i
n
k
.
s
i
n
[
(
n
+
1
)
k
−
(
n
k
)
]
c
o
s
(
n
+
1
)
k
.
c
o
s
n
k
⇒
T
n
=
1
s
i
n
k
[
t
a
n
(
n
+
1
)
k
−
t
a
n
(
n
k
)
]
so,
∑
88
n
=
0
T
n
=
1
s
i
n
k
[
t
a
n
89
k
−
t
a
n
0
]
=
c
o
s
k
.
c
o
s
e
c
2
k
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Similar Questions
Q1
Let
K
=
1
∘
, then prove that
88
∑
n
=
0
1
cos
n
K
.
cos
(
n
+
1
)
K
=
cos
K
sin
2
K
View Solution
Q2
The value of
∑
88
n
=
0
1
c
o
s
n
k
.
c
o
s
(
n
+
1
)
k
, where k is
1
o
is equal to
View Solution
Q3
Let
F
(
k
)
=
(
1
+
sin
π
2
k
)
(
1
+
sin
(
k
−
1
)
π
2
k
)
(
1
+
sin
(
2
k
+
1
)
π
2
k
)
(
1
+
sin
(
3
k
−
1
)
π
2
k
)
.
The value of
F
(
1
)
+
F
(
2
)
+
F
(
3
)
is equal to
View Solution
Q4
The value of
∑
n
+
1
r
=
1
(
∑
n
k
=
1
k
C
r
−
1
)
where r, k, n
ϵ
N is equal to
View Solution
Q5
If for
k
∈
N
,
sin
2
k
x
sin
x
=
2
[
cos
x
+
cos
3
x
+
.
.
.
+
cos
(
2
k
−
1
)
x
]
,
then value of
I
=
∫
π
/
2
0
sin
2
k
x
cot
x
d
x
is
View Solution