Question

If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot C are in
(a) GP
(b) HP
(c) AP
(d) None of these

Answer 1

(b) HP

Given:
sin (B + C − A), sin (C + A − B) and sin (A + B − C) are in A.P.

$$\begin{gathered} \Rightarrow \sin \left(C+A-B\right)-\sin \left(B+C-A\right)=\sin \left(A+B-C\right)-\sin \left(C+A-B\right) \\ \Rightarrow 2\sin \left(\frac{C+A-B-B-C+A}{2}\right)\cos \left(\frac{C+A-B+B+C-A}{2}\right)=2\sin \left(\frac{A+B-C-C-A+B}{2}\right)\cos \left(\frac{A+B-C+C+A-B}{2}\right) \\ \Rightarrow \sin \left(A-B\right)\cos C=\sin \left(B-C\right)\cos A \\ \Rightarrow \sin A\cos B\cos C-\cos A\sin B\cos C=\sin B\cos C\cos A-\cos B\sin C\cos A \\ \Rightarrow 2\sin B\cos A\cos C=\sin A\cos B\cos C+\cos A\cos B\sin C \\ \text{Dividing both sides by cos AcosBcosC:} \\ 2\tan B=\tan A+\tan C \\ \Rightarrow \frac{2}{\mathrm{co}tB}=\frac{1}{cotA}+\frac{1}{cotC} \end{gathered}$$

Hence, cotA, cotB and cotC are in HP.