Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If Sin 4 α + 4 cos4 β + 2 = 4√2 sin α cosβ ; α , β ∈ [0, π] , then cos (α + β) - cos(α - β)is equal to:
Q. If Sin
4
α
+
4
co
s
4
β
+
2
=
4
2
s
in
α
cos
β
;
α
,
β
∈
[
0
,
π
]
,
then cos
(
α
+
β
)
−
cos
(
α
−
β
)
is equal to:
4339
227
JEE Main
JEE Main 2019
Trigonometric Functions
Report Error
A
0
25%
B
−
2
34%
C
−
1
21%
D
2
19%
Solution:
A
.
M
.
≥
G
.
M
.
4
s
i
n
4
α
+
4
c
o
s
4
β
+
1
+
1
≥
(
sin
4
α
.4
cos
4
β
.1.1
)
4
1
sin
4
+
4
cos
2
β
+
2
≥
4
2
sin
α
cos
β
given that
sin
4
α
+
4
cos
4
β
+
2
=
4
2
sin
α
cos
β
⇒
A
.
M
.
=
G
.
M
.
⇒
sin
4
α
=
1
=
4
cos
4
β
sin
α
=
1
,
cos
β
=
±
;
2
1
⇒
sin
β
=
2
1
a
s
β
∈
[
0
,
π
]
cos
(
α
+
β
)
−
cos
(
α
−
β
)
=
−
2
sin
α
sin
β
=
−
2