If a_1 ,a_2 ,a_3 are in AP and d is the common difference then tan^(−1) ((d/(1+a_1 a_2 )))+tan^(−1) ((d/(1+a_2 a


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Question Number 130581 by bramlexs22 last updated on 27/Jan/21

$If a_{1}, a_{2}, a_{3} are in AP and d is the common$ $difference then \tan^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{d}{1 + a_{2} a_{3}}) = ?$
	Answered by som(math1967) last updated on 27/Jan/21

	$\tan^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{2} - a_{3}})$ $= \tan^{- 1} a_{2} - \tan^{- 1} a_{1} + \tan^{- 1} a_{3} - \tan^{- 1} a_{2}$ $= \tan^{- 1} a_{3} - \tan^{- 1} a_{1}$ $[d = a_{2} - a_{1} = a_{3} - a_{2}]$
	Commented by bramlexs22 last updated on 27/Jan/21

	Commented by bramlexs22 last updated on 27/Jan/21

	$= \tan^{- 1} (\frac{a_{3} - a_{1}}{1 + a_{1} a_{3}}) = \tan^{- 1} (\frac{2 d}{1 + a_{1} a_{3}})$
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