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Question
Mathematics
The value of cot(∑ limits19n=1 cot-1 (1+ ∑ limitsnp=1 2p)) is :
Q. The value of
cot
(
n
=
1
∑
19
cot
−
1
(
1
+
p
=
1
∑
n
2
p
)
)
is :
8258
214
JEE Main
JEE Main 2019
Inverse Trigonometric Functions
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A
23
22
12%
B
22
23
28%
C
19
21
50%
D
21
19
10%
Solution:
cot
(
∑
n
=
1
19
cot
−
1
(
1
+
n
(
n
+
1
)
)
)
cot
(
∑
n
=
1
19
cot
−
1
(
n
2
+
n
+
1
)
)
=
cot
(
∑
n
=
1
19
tan
−
1
1
+
n
(
n
+
1
)
1
)
∑
n
=
1
19
(
tan
−
1
(
n
+
1
)
−
tan
−
1
n
)
cot
(
tan
−
1
20
−
tan
−
1
1
)
=
c
o
t
β
−
c
o
t
A
c
o
t
A
c
o
t
β
+
1
(Where
tan
A
=
20
,
tan
B
=
1
)
1
−
20
1
1
(
20
1
)
+
1
=
19
21