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If A+B+C=pi Show that tanA=tanB+tanC If (cot(alpha-beta)/ cot alpha ) +(cos^2gama/cos^2 alpha)=1
3 tanФ = Cot Ф = 1/tanФ3 tan²Ф = 1TanФ = + 1/√3 Ф = n π + π/6Ф = π/6 or π+π/6 = 7π/6 or π-π/6 = 5π/6 or 2π-π/6 = 11π/6 There are 4 solutions for 0° =============== A + B + C = π/2 => Tan(A+B) = Tan(π/2-C) = Cot C = 1/TanC => Tan (B+C) = 1/TanA => Tan(C+A) = 1/TanB Tan (A+B) = [TanA + TanB]/[1 - TanA TanB]Rearrange: TanA TanB = 1 - (TanA+TanB)/Tan(A+B) = 1 - (TanA+TAnB) ×TanC ---(1) Similarly: Tan B TanC = 1 - (TanB + TanC) × TanA ---(2) Tan C TanA = 1 - (TanC + TanA) ×TanB --- (3) Add (1) , (2) & (3):LHS = TanA TanB + TanB TanC + TanC Tan A = 3 - 2 (TanA TanB + TanB TanC + Tanc TanA) = 3 - 2 LHS So TanA TanB + TanB TanC + Tan C TanA = 1
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