cot 105° = cot (60 +45)
cotangente de la suma de dos angulos
cot (α+β) = (cotα .cotβ - 1)/cotβ +cotα
reemplazamos
cot (60 +45)
cot (60 +45) = (cot60 .cot45 - 1)/cot45 +cot60
cot (105) = (1/√3 .1 - 1)/(1 +1/√3)
cot 105 = (1/√3 -1)/(1 +1/√3)
cot 105 = (1/√3 -1)²/(1/√3 +1)(1/√3 -1)
cot 105 = (1/√3 -1)²/(1/3 -1)
cot 105 = (1/3 -2/√3 + 1)/-2/3
cot 105 = (4/3 -2/√3)/-2/3
cot 105 = - 3(4/3 -2/√3)/2
cot 105 = - 3(4/3 -2√3/3)/2
cot 105 = - (4 -2√3)/2
cot 105 = - (2 -√3)
cot 105 = - 2+√3
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cot 105° = cot (60 +45)
cotangente de la suma de dos angulos
cot (α+β) = (cotα .cotβ - 1)/cotβ +cotα
reemplazamos
cot (60 +45)
cot (60 +45) = (cot60 .cot45 - 1)/cot45 +cot60
cot (105) = (1/√3 .1 - 1)/(1 +1/√3)
cot 105 = (1/√3 -1)/(1 +1/√3)
cot 105 = (1/√3 -1)²/(1/√3 +1)(1/√3 -1)
cot 105 = (1/√3 -1)²/(1/3 -1)
cot 105 = (1/3 -2/√3 + 1)/-2/3
cot 105 = (4/3 -2/√3)/-2/3
cot 105 = - 3(4/3 -2/√3)/2
cot 105 = - 3(4/3 -2√3/3)/2
cot 105 = - (4 -2√3)/2
cot 105 = - (2 -√3)
cot 105 = - 2+√3