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Byju's Answer
Standard XII
Mathematics
Domain and Range of Trigonometric Ratios
Show that s...
Question
Show that
s
i
n
4
θ
−
c
o
s
4
θ
=
1
−
2
c
o
s
2
θ
Open in App
Solution
LHS
=
s
i
n
4
θ
−
c
o
s
4
θ
=
(
s
i
n
2
θ
)
2
−
(
c
o
s
2
θ
)
2
[
∵
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
]
=
(
s
i
n
2
θ
+
c
o
s
2
θ
)
(
s
i
n
2
θ
−
c
o
s
2
θ
)
=
s
i
n
2
θ
−
c
o
s
2
θ
[
∵
s
i
n
2
θ
+
c
o
s
2
θ
=
1
]
=
−
(
c
o
s
2
θ
−
s
i
n
2
θ
)
[
∵
c
o
s
2
θ
=
c
o
s
2
θ
−
s
i
n
2
=
2
c
o
s
2
θ
−
1
]
=
−
c
o
s
2
θ
=
−
(
2
c
o
s
2
θ
−
1
)
=
1
−
2
c
o
s
2
θ
=
R
.
H
.
S
.
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Similar questions
Q.
sin
4
θ – cos
4
θ = 1 – 2cos
2
θ
Q.
Find the value
s
i
n
4
θ
-
c
o
s
4
θ
+ 2
c
o
s
2
θ
, when
θ
=
Π
7
__
Q.
prove that
cos
4
θ
−
sin
4
θ
=
1
−
2
sin
2
θ
Q.
Prove:
sin
4
θ
−
cos
4
θ
=
1
−
2
cos
2
θ
Q.
Solve:
s
i
n
4
θ
+
2
s
i
n
2
θ
c
o
s
2
θ
+
c
o
s
4
θ
=
1
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