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Tardigrade
Question
Mathematics
| sin 2 θ cos 2 θ cos 2 θ sin 2 θ|=
Q.
∣
∣
sin
2
θ
cos
2
θ
cos
2
θ
sin
2
θ
∣
∣
=
1584
200
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Gujarat CET 2019
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A
cos
2
θ
B
2
1
(
1
+
cos
2
2
θ
)
C
2
1
(
1
−
sin
2
2
θ
)
D
2
1
sin
2
2
θ
Solution:
sin
4
θ
+
cos
4
θ
=
1
−
2
sin
2
θ
cos
2
θ
=
1
−
2
1
sin
2
2
θ
=
1
−
2
1
(
1
−
cos
2
2
θ
)
=
2
1
(
1
+
cos
2
2
θ
)