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Problem
Problem
In a triangle ABC, if
sin
A
sin
(
B
−
C
)
=
sin
C
sin
(
A
−
B
)
\sin A \sin(B-C) = \sin C \sin(A-B)
sin
A
sin
(
B
−
C
)
=
sin
C
sin
(
A
−
B
)
, then prove that
cot
A
,
cot
B
,
cot
C
\cot A, \cot B, \cot C
cot
A
,
cot
B
,
cot
C
are in AP.
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