If A,B,C are angles of a triangle show that tan^(−1) (cot Acot B)+tan^(−1) (cot Bcot C)+tan^(−1) (cot Ccot A) = tan^(−1) {1+((8cos Acos Bcos C)/(sin^2 2A+sin^2 2B+sin^2 2C))}


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Question Number 54536 by 951172235v last updated on 05/Feb/19
$If A,B,C are angles of a triangle show that tan^(−1) (cot Acot B)+tan^(−1) (cot Bcot C)+tan^(−1) (cot Ccot A) = tan^(−1) {1+((8cos Acos Bcos C)/(sin^2 2A+sin^2 2B+sin^2 2C))}$
$If A, B, C are angles of a triangle show that$ $\tan^{- 1} (\cot Acot B) + \tan^{- 1} (\cot Bcot C) + \tan^{- 1} (\cot Ccot A)$ $= \tan^{- 1} {1 + \frac{8 \cos Acos Bcos C}{\sin^{2} 2 A + \sin^{2} 2 B + \sin^{2} 2 C}}$
	Answered by 951172235v last updated on 08/Feb/19
	$A+B+C =Λ^− tan A+tan B+tan C = tan Atan Btan tan C tan α =cot Acot B tan β =cot Bcot C tan γ =cot Ccot A tan α+tan β+tan γ =1 tan (α+β+γ) =((Σtan α−tan αtan βtan γ)/(1−Σtan αtan β)) = ((1−(cot Acot Bcot C)^2 )/(1−cot Acot Bcot C(cot A+cot B+cot C))) =((tan Atan Btan C−cot Acot Bcot C)/(Σtan A−Σcot A)) = ((tan Atan Btan C−cot Acot Bcot C)/(Σ(((−cos 2A)/(sin 2A))))) = ((8(sin Asin Bsin C)^2 −8(cos Acos Bcos C)^2 )/(−Σsin 2A(sin 2Bcos 2C+sin 2Ccos 2B))) = (((1/2){(Σsin 2A)^2 −[1+Σcos 2A]^2 })/(Σsin^2 2A)) =(((1/2){2Σsin^2 A −4Σcos 2A−4})/(Σsin^2 2A)) = ((Σsin^2 2A+8cos Acos Bcos C)/(Σsin^2 2A)) =1+ ((8cos Acos Bcos C)/(Σsin^2 2A)) α+β+γ =tan^(−1) {1+((8cos Acos Bcos C)/(Σsin^2 2A))} ans.$
	$A + B + C = \bar{Λ}$ $\tan A + \tan B + \tan C = \tan Atan Btan \tan C$ $\tan α = \cot Acot B \tan β = \cot Bcot C \tan γ = \cot Ccot A$ $\tan α + \tan β + \tan γ = 1$ $\tan (α + β + γ) = \frac{Σ \tan α - \tan α \tan β \tan γ}{1 - Σ \tan α \tan β}$ $= \frac{1 - {(\cot Acot Bcot C)}^{2}}{1 - \cot Acot Bcot C (\cot A + \cot B + \cot C)}$ $= \frac{\tan Atan Btan C - \cot Acot Bcot C}{Σ \tan A - Σ \cot A}$ $= \frac{\tan Atan Btan C - \cot Acot Bcot C}{Σ (\frac{- \cos 2 A}{\sin 2 A})}$ $= \frac{8 {(\sin Asin Bsin C)}^{2} - 8 {(\cos Acos Bcos C)}^{2}}{- Σ \sin 2 A (\sin 2 Bcos 2 C + \sin 2 Ccos 2 B)}$ $= \frac{\frac{1}{2} {{(Σ \sin 2 A)}^{2} - {[1 + Σ \cos 2 A]}^{2}}}{Σ \sin^{2} 2 A}$ $= \frac{\frac{1}{2} {2 Σ \sin^{2} A - 4 Σ \cos 2 A - 4}}{Σ \sin^{2} 2 A}$ $= \frac{Σ \sin^{2} 2 A + 8 \cos Acos Bcos C}{Σ \sin^{2} 2 A}$ $= 1 + \frac{8 \cos Acos Bcos C}{Σ \sin^{2} 2 A}$ $α + β + γ = \tan^{- 1} {1 + \frac{8 \cos Acos Bcos C}{Σ \sin^{2} 2 A}} ans .$
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