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Byju's Answer
Standard VI
Biology
Pond Habitat
Question 4sin...
Question
Question 4
(
s
i
n
α
+
c
o
s
α
)
(
t
a
n
α
+
c
o
t
α
)
=
s
e
c
α
+
c
o
s
e
c
α
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Solution
L
H
S
=
(
s
i
n
α
+
c
o
s
α
)
(
t
a
n
α
+
c
o
t
α
)
=
(
s
i
n
α
+
c
o
s
α
)
(
s
i
n
α
c
o
s
α
+
c
o
s
α
s
i
n
α
)
[
∵
t
a
n
θ
=
s
i
n
θ
c
o
s
θ
a
n
d
c
o
t
θ
=
c
o
s
θ
s
i
n
θ
]
=
(
s
i
n
α
+
c
o
s
α
)
(
s
i
n
2
α
+
c
o
s
2
α
s
i
n
α
.
c
o
s
α
)
[
∵
s
i
n
2
θ
+
c
o
s
2
θ
=
1
]
=
1
c
o
s
α
+
1
s
i
n
α
[
∵
s
e
c
θ
=
1
c
o
s
θ
a
n
d
c
o
s
e
c
θ
=
1
s
i
n
θ
]
=
s
e
c
α
+
c
o
s
e
c
α
Suggest Corrections
3
Similar questions
Q.
Question 6
1
+
c
o
t
2
α
1
+
c
o
s
e
c
α
=
c
o
s
e
c
α
Q.
if
cot
α
+
tan
α
=
m
and
1
cos
α
−
cos
α
=
n
then eliminate
α
.
Q.
For
α
being acute solve:
cos
α
=
cot
α
.
Q.
The values of
c
o
s
α
,
t
a
n
α
,
c
o
t
α
,
respectively if
s
i
n
α
=
5
13
and
π
2
<
α
<
π
Q.
If
s
e
c
α
=
2
√
3
then find the value of
1
−
c
o
s
e
c
α
1
+
c
o
s
e
c
α
where
α
is in IV quadrant.
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