If \( \sin(B+C-A), \sin(C+A-B), \sin(A+B-C) \) are in A.P., then \( \cot A, \cot B, \cot C \) are in
Options:
(a) GP
(b) HP
(c) AP
(d) none of these
Solution:
Since the given terms are in A.P.,
\[
2\sin(C+A-B)
=
\sin(B+C-A)+\sin(A+B-C)
\]
Using identity,
\[
\sin x+\sin y
=
2\sin\frac{x+y}{2}\cos\frac{x-y}{2}
\]
\[
2\sin(C+A-B)
=
2\sin C \cos(A-B)
\]
\[
\sin(C+A-B)
=
\sin C \cos(A-B)
\]
Using,
\[
\sin(C+A-B)
=
\sin C\cos(A-B)+\cos C\sin(A-B)
\]
Therefore,
\[
\sin C\cos(A-B)+\cos C\sin(A-B)
=
\sin C\cos(A-B)
\]
\[
\cos C\sin(A-B)=0
\]
Hence,
\[
\sin(A-B)=0
\]
\[
A=B
\]
Therefore,
\[
\cot A=\cot B
\]
So,
\[
\cot A,\cot B,\cot C
\]
are in A.P.
\[
\boxed{\text{AP}}
\]
Correct option: (c)