Mark the Correct alternative in the following:

If sin (B + C – A), sin (C + A – B), sin (A + B – C) are in A.P., then cot A, cot B, cot C are in


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


⇒ 2sin (C + A – B) = sin (B + C – A) + sin (A + B – C)


⇒ 2sin (C + A – B) = 2sin {(B + C – A + A + B – C)/2} cos {(B + C – A – A – B + C)/2}


⇒ sin {(C + A) – B} = sin B cos (C – A)


⇒ sin (C + A) cos B – cos (C + A) sin B = sin B cos (C – A)


⇒ {sin C cos A + cos C sin A} cos B – {cos C cos A – sin C sin A} sin B = sin B {cos C cos A + sin C sin A}


⇒ cos A cos B sin C + sin A cos B cos C – cos A sin B cos C + sin A sin B sin C = cos A sin B cos C + sin A sin B sin C


⇒ cos A cos B sin C + sin A cos B cos C = 2cos A sin B cos C


⇒ cos B (cos A sin C + sin A cos C) = 2cos A sin B cos C





∴ cot A, cot B, cot C are in H.P. (Ans)

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