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-

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 + \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 = \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 .