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WBJEE 2024 Math Paper
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© examsnet.com
Question : 56
Total: 75
If
α
1
,
α
2
,
.
.
.
,
α
n
are in A.P. with common difference
θ
, then the sum of the series
s
e
c
α
1
s
e
c
α
2
+
s
e
c
α
2
s
e
c
α
3
+
.
.
.
+
s
e
c
α
n
−
1
s
e
c
α
n
=
k
(
tan
α
n
−
tan
α
1
)
where
k
=
sin
θ
cos
θ
sec
θ
cosec
θ
Validate
Solution:
α
2
−
α
1
=
α
3
−
α
2
=
.
.
.
.
.
.
=
α
n
−
α
n
−
1
=
θ
sec
α
1
sec
α
2
+
sec
α
2
sec
α
3
+
.
.
.
+
sec
α
n
−
1
sec
α
n
=
1
cos
α
1
cos
α
2
+
1
cos
α
2
cos
α
3
+
.
.
.
.
.
.
+
1
cos
α
n
−
1
cos
α
n
=
1
sin
θ
[
sin
θ
cos
α
1
cos
α
2
+
sin
θ
cos
α
2
cos
α
3
+
.
.
.
+
+
sin
θ
cos
α
n
−
1
cos
α
n
]
=
1
sin
θ
[
sin
(
α
2
−
α
1
)
cos
α
1
cos
α
2
+
sin
(
α
3
−
α
2
)
cos
α
2
cos
α
3
+
.
.
.
+
sin
(
α
n
−
α
n
−
1
)
cos
α
n
−
1
cos
α
n
]
=
1
sin
θ
[
(
tan
α
2
−
tan
α
1
)
+
(
tan
α
3
−
tan
α
2
)
+
.
.
.
+
(
tan
α
n
−
tan
α
n
−
1
)
]
=
1
sin
θ
[
tan
α
n
−
tan
α
1
]
=
cosec
θ
(
tan
α
n
−
tan
α
1
)
=
k
(
tan
α
n
−
tan
α
1
)
Comparing we get
k
=
cosec
θ
© examsnet.com
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