If \( A, B, C \) are in A.P., then \( \dfrac{\sin A-\sin C}{\cos C-\cos A} \) is
Options:
(a) \( \tan B \)
(b) \( \cot B \)
(c) \( \tan2B \)
(d) none of these
Solution:
Since \( A, B, C \) are in A.P.,
\[
2B=A+C
\]
Using identity,
\[
\sin A-\sin C
=
2\cos\frac{A+C}{2}\sin\frac{A-C}{2}
\]
Also,
\[
\cos C-\cos A
=
2\sin\frac{A+C}{2}\sin\frac{A-C}{2}
\]
Therefore,
\[
\frac{\sin A-\sin C}{\cos C-\cos A}
=
\frac{
2\cos\frac{A+C}{2}\sin\frac{A-C}{2}
}{
2\sin\frac{A+C}{2}\sin\frac{A-C}{2}
}
\]
\[
=
\cot\frac{A+C}{2}
\]
Since,
\[
A+C=2B
\]
\[
=
\cot B
\]
\[
\boxed{\cot B}
\]
Correct option: (b)