Question

If A, B, C are in A.P., then $\frac{\sin A-\sin C}{\cos C-\cos A}$=
(a) tan B
(b) cot B
(c) tan 2 B
(d) None of these

Answer 1

(b) cot B

Since A,B and C are in A.P,
$$\begin{gathered} B-A=C-B \\ \mathrm{or},2B=A+C \end{gathered}$$

$$\begin{gathered} \frac{\sin A-\sin C}{\cos C-\cos A} \\ =\frac{2\sin \left(\frac{A-C}{2}\right)\cos \left(\frac{A+C}{2}\right)}{-2\sin \left(\frac{C+A}{2}\right)\sin \left(\frac{C-A}{2}\right)}\left[\because \sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\mathrm{and}\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right] \\ =\frac{\sin \left(\frac{A-C}{2}\right)\cos \left(\frac{A+C}{2}\right)}{-\sin \left(\frac{A+C}{2}\right)\sin \left(\frac{C-A}{2}\right)} \end{gathered}$$

$$\begin{gathered} =\frac{\sin \left(\frac{A-C}{2}\right)\cos \left(\frac{A+C}{2}\right)}{\sin \left(\frac{A+C}{2}\right)\sin \left(\frac{A-C}{2}\right)} \\ =\frac{\cos \left(\frac{A+C}{2}\right)}{\sin \left(\frac{A+C}{2}\right)} \\ =\frac{\cos B}{\sin B} \\ =\cot B \end{gathered}$$