If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in
Options
GP
HP
AP
None of these
Solution
HP
Given:
sin (B + C − A), sin (C + A − B) and sin (A + B − C) are in A.P.
\[\Rightarrow \sin\left( C + A - B \right) - \sin\left( B + C - A \right) = \sin\left( A + B - C \right) - \sin\left( C + A - B \right)\]
\[ \Rightarrow 2\sin\left( \frac{C + A - B - B - C + A}{2} \right) \cos\left( \frac{C + A - B + B + C - A}{2} \right) = 2\sin\left( \frac{A + B - C - C - A + B}{2} \right) \cos\left( \frac{A + B - C + C + A - B}{2} \right)\]
\[ \Rightarrow \sin\left( A - B \right) \cos C = \sin\left( B - C \right) \cos A\]
\[ \Rightarrow \sin A \cos B \cos C - \cos A \sin B \cos C = \sin B \cos C\cos A - \cos B \sin C \cos A\]
\[ \Rightarrow 2\sin B \cos A \cos C = \sin A \cos B \cos C + \cos A \cos B \sin C\]
Dividing both sides by cosA cosB cosC:
\[2\tan B = \tan A + \tan C \]
\[ \Rightarrow \frac{2}{cotB} = \frac{1}{cotA} + \frac{1}{cotC}\]
Hence, cotA, cotB and cotC are in HP.
