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JEE Main 17 March 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 65
Total: 90
If
cot
−
1
(
α
)
=
cot
−
1
2
+
cot
−
1
8
+
cot
−
1
18
+
cot
−
1
32
+
.
.
.
.
.
upto 100 terms, then
α
is
[17 Mar 2021 Shift 1]
1.01
1.00
1.02
1.03
Validate
Solution:
cot
−
1
α
=
cot
−
1
2
+
cot
−
1
8
+
cot
−
1
18
+
cot
−
1
32
.
.
.
upto 100 terms
Let
T
n
be the
n
th term of
cot
−
1
α
⇒
T
n
=
cot
−
1
(
2
n
2
)
=
tan
−
1
(
1
2
n
2
)
=
tan
−
1
[
(
2
n
+
1
)
−
(
2
n
−
1
)
1
+
(
2
n
+
1
)
(
2
n
−
1
)
]
∵
tan
−
1
x
−
tan
−
1
y
=
tan
−
1
(
x
−
y
1
+
x
y
)
∴
tan
−
1
{
(
2
n
+
1
)
−
(
2
n
−
1
)
1
+
(
2
n
+
1
)
(
2
n
−
1
)
}
=
tan
−
1
(
2
n
+
1
)
−
tan
−
1
(
2
n
−
1
)
T
1
=
tan
−
1
3
−
tan
−
1
1
T
2
=
tan
−
1
5
−
tan
−
1
3
T
3
=
tan
−
1
7
−
tan
−
1
5
:
T
99
=
tan
−
1
199
−
tan
−
1
197
T
100
=
tan
−
1
201
−
tan
−
1
199
100
∑
n
=
1
T
r
=
tan
−
1
201
−
tan
−
1
1
=
tan
−
1
(
201
−
1
1
+
201.1
)
=
tan
−
1
(
200
202
)
⇒
cot
−
1
(
α
)
=
tan
−
1
(
200
202
)
⇒
cot
−
1
(
α
)
=
cot
−
1
(
202
200
)
∴
α
=
202
200
=
101
100
=
1.01
© examsnet.com
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