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Byju's Answer
Standard VI
Biology
Pond Habitat
Question 15Sh...
Question
Question 15
Show that:
t
a
n
4
θ
+
t
a
n
2
θ
=
s
e
c
4
θ
−
s
e
c
2
θ
Open in App
Solution
L
H
S
=
t
a
n
4
θ
+
t
a
n
2
θ
=
t
a
n
2
θ
(
t
a
n
2
θ
+
1
)
=
t
a
n
2
θ
.
s
e
c
2
θ
[
∵
s
e
c
2
θ
=
t
a
n
2
θ
+
1
]
=
(
s
e
c
2
θ
−
1
)
s
e
c
2
θ
[
∵
t
a
n
2
θ
=
s
e
c
2
θ
−
1
]
=
s
e
c
4
θ
−
s
e
c
2
θ
=
R
H
S
Suggest Corrections
92
Similar questions
Q.
Prove that
sec
4
θ
−
sec
2
θ
=
tan
4
θ
+
tan
2
θ
Q.
Question 2
Prove that:
√
s
e
c
2
θ
+
c
o
s
e
c
2
θ
=
t
a
n
θ
+
c
o
t
θ
Q.
Prove that
(
i
)
s
e
c
2
θ
+
c
o
s
e
c
2
θ
=
s
e
c
2
θ
c
o
s
e
c
2
θ
(
i
i
)
t
a
n
2
θ
−
s
i
n
2
θ
=
t
a
n
2
θ
s
i
n
2
θ
(
i
i
i
)
t
a
n
2
θ
+
c
o
t
2
θ
+
2
=
s
e
c
2
θ
c
o
s
e
c
2
θ
[3 MARKS]
Q.
Prove that :
2
sec
2
θ
−
sec
4
θ
−
c
o
sec
2
θ
−
c
o
sec
4
θ
=
cot
4
θ
−
tan
4
θ
Q.
If
θ
is any angle, then
s
e
c
4
θ
−
s
e
c
2
θ
is equal to
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