✨ ベストアンサー ✨
x = (√5-1)/2 時，x²+2(sinθ+cosθ)x+1=0 成立
此時 sinθ+cosθ = -(1+x²)/2x
tanθ + 1/tanθ
= sinθ/cosθ + cosθ/sinθ
= (sin²θ+cos²θ)/(sinθcosθ)
= 1/sinθcosθ
透過對 sinθ+cosθ 平方即可產生 sinθcosθ
(sinθ+cosθ)² = [ -(1+x²)/2x ]²
sin²θ+cos²θ+2sinθcosθ = (1+x²)²/4x²
1+2sinθcosθ = (1+2x²+x⁴)/4x²
2sinθcosθ = (1-2x²+x⁴)/4x²
= (1-x²)²/4x²
sinθcosθ = (1-x²)²/8x²
1/sinθcosθ = 8x²/(1-x²)²
其中，x² = (6-2√5)/4 = (3-√5)/2
1/sinθcosθ = 4(3-√5) / ((√5-1)/2)²
= 16 (3-√5) / (√5-1)²
= 16 (3-√5) / (6-2√5)
= 8 (3-√5) / (3-√5)
= 8
所以 tanθ + 1/tanθ = 8
謝謝你！