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If
{
sin
(
α
−
β
)
+
cos
(
α
+
2
β
)
sin
β
}
2
=
4
cos
α
sin
β
sin
(
α
+
β
)
\{\sin(\alpha - \beta) + \cos(\alpha + 2\beta)\sin\beta\}^2 = 4 \cos \alpha \sin \beta \sin(\alpha + \beta)
{
sin
(
α
−
β
)
+
cos
(
α
+
2
β
)
sin
β
}
2
=
4
cos
α
sin
β
sin
(
α
+
β
)
…
Mathematics
Problem
If
{
sin
(
α
−
β
)
+
cos
(
α
+
2
β
)
sin
β
}
2
=
4
cos
α
sin
β
sin
(
α
+
β
)
\{\sin(\alpha - \beta) + \cos(\alpha + 2\beta)\sin\beta\}^2 = 4 \cos \alpha \sin \beta \sin(\alpha + \beta)
{
sin
(
α
−
β
)
+
cos
(
α
+
2
β
)
sin
β
}
2
=
4
cos
α
sin
β
sin
(
α
+
β
)
. Then, prove that
tan
α
+
tan
β
=
tan
β
(
2
cos
β
−
1
)
2
;
\tan \alpha + \tan \beta = \frac{\tan \beta}{(\sqrt{2} \cos \beta - 1)^2};
tan
α
+
tan
β
=
(
2
c
o
s
β
−
1
)
2
t
a
n
β
;
α
,
β
∈
(
0
,
π
/
4
)
\alpha, \beta \in (0, \pi/4)
α
,
β
∈
(
0
,
π
/4
)
.
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