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Test Index
JEE Mains 27-July-2022 Shift 2 Solved Paper
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© examsnet.com
Question : 19
Total: 90
Let
S
=
{
θ
∈
(
0
,
π
2
)
:
∑
m
=
1
9
sec
(
θ
+
(
m
−
1
)
π
6
)
sec
(
θ
+
m
π
6
)
=
−
8
√
3
}
. Then
[27-Jul-2022-Shift-2]
S
=
{
π
12
}
S
=
{
2
π
3
}
∑
θ
∈
S
θ
=
π
2
∑
θ
∈
S
θ
=
3
π
4
Validate
Solution:
s
=
{
0
∈
(
0
,
π
2
)
:
9
∑
m
=
1
s
e
c
(
θ
+
(
m
−
1
)
π
6
)
s
e
c
(
θ
+
m
π
6
)
=
−
8
√
3
}
.
9
∑
m
=
1
1
cos
(
θ
+
(
m
−
1
)
π
6
)
cos
(
θ
+
m
π
6
)
1
s
i
n
(
π
6
)
9
∑
m
=
1
s
i
n
[
(
θ
+
m
π
6
)
−
(
θ
+
(
m
−
1
)
π
6
)
]
cos
(
θ
+
(
m
−
1
)
π
6
)
cos
(
θ
+
π
6
)
=
2
9
∑
m
=
1
[
tan
(
θ
+
m
π
6
)
−
tan
(
θ
+
(
m
−
1
)
π
6
)
]
Now,
m
=
1
2
[
tan
(
θ
+
π
6
)
−
tan
(
θ
)
]
m
=
2
2
[
tan
(
θ
+
2
π
6
)
−
tan
(
θ
+
π
6
)
]
.
.
.
m
=
9
2
[
tan
(
θ
+
9
π
6
)
−
tan
(
θ
+
8
π
6
)
]
∴
=
2
[
tan
(
θ
+
3
π
2
)
−
tan
θ
]
=
−
8
√
3
=
−
2
[
cot
θ
+
tan
θ
]
=
−
8
√
3
=
−
2
×
2
2
s
i
n
θ
cos
θ
=
−
8
√
3
=
1
s
i
n
2
θ
=
2
√
3
⇒
s
i
n
2
θ
=
√
3
2
2
θ
=
π
3
2
θ
=
2
π
3
θ
=
π
6
θ
=
π
3
∑
θ
i
=
π
6
+
π
3
=
π
2
© examsnet.com
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