tanB/2=4tanA/2 then proved tan(B-A/2)=3sinA/5-3cosA
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Answer:
Given
tan
(
β
2
)
=
4
tan
(
α
2
)
then
tan
(
β
−
α
2
)
=
tan
(
β
2
)
−
tan
(
α
2
)
1
+
tan
(
β
2
)
tan
(
α
2
)
=
4
tan
(
α
2
)
−
tan
(
α
2
)
1
+
4
tan
(
α
2
)
tan
(
α
2
)
=
3
tan
(
α
2
)
1
+
4
tan
2
(
α
2
)
=
3
sin
(
α
2
)
cos
(
α
2
)
1
+
4
sin
2
(
α
2
)
cos
2
(
α
2
)
=
3
⋅
2
sin
(
α
2
)
⋅
cos
(
α
2
)
2
cos
2
(
α
2
)
+
8
sin
2
(
α
2
)
=
3
sin
α
1
+
cos
α
+
4
(
1
−
cos
α
)
=
3
sin
α
5
−
3
cos
α
=
R
H
S
Step-by-step explanation:
please aad branilest
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