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If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P.,

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

Solution:

(b) HP

Given:

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

$\Rightarrow \sin (C+A-B)-\sin (B+C-A)=\sin (A+B-C)-\sin (C+A-B)$

$\Rightarrow 2 \sin \left(\frac{C+A-B-B-C+A}{2}\right) \cos \left(\frac{C+A-B+B+C-A}{2}\right)=2 \sin \left(\frac{A+B-C-C-A+B}{2}\right) \cos \left(\frac{A+B-C+C+A-B}{2}\right)$

$\Rightarrow \sin (A-B) \cos C=\sin (B-C) \cos A$

$\Rightarrow \sin A \cos B \cos C-\cos A \sin B \cos C=\sin B \cos C \cos A-\cos B \sin C \cos A$

$\Rightarrow 2 \sin B \cos A \cos C=\sin A \cos B \cos C+\cos A \cos B \sin C$

Dividing both sides by $\cos A \cos B \cos C$ :

$2 \tan B=\tan A+\tan C$

$\Rightarrow \frac{2}{\cot B}=\frac{1}{\cot A}+\frac{1}{\cot C}$

Hence, cotA, cotB and cotC are in HP.

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