三角函數的題目

1.
[1 / (1 + sinθ)] + [1 / (1 - sinθ)]
= [(1 - sinθ) / (1 + sinθ)(1 - sinθ)] + [(1 + sinθ)/(1 + sinθ)(1 - sinθ)]
= [(1 - sinθ) / (1 - sin²θ)] + [(1 + sinθ) / (1 - sin²θ)]
= [(1 - sinθ) + (1 + sinθ)] / (1-sin²θ)
= 2 / sinθ


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2.
[1 / (1 + sinθ)] + [1 / (1 - sinθ)]
= [(1 - sinθ) / (1 + sinθ)(1 - sinθ)] + [(1 + sinθ)/(1 + sinθ)(1 - sinθ)]
= [(1 - sinθ) / (1 - sin²θ)] + [(1 + sinθ) / (1 - sin²θ)]
= [(1 - sinθ) + (1 + sinθ)] / (1-sin²θ)
= 2 / cos²θ
= 2 / sinθ


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3.
2 sinθ + cosθ = 2
2[2sin(θ/2)cos(θ/2)] + [cos²(θ/2) - sin²(θ/2)]= 2[sin²(θ/2) + cos²(θ/2)]
4sin(θ/2)cos(θ/2) + cos²(θ/2) - sin²(θ/2)= 2sin²(θ/2) + 2cos²(θ/2)
3sin²(θ/2) - 4sin(θ/2)cos(θ/2) + cos²(θ/2) = 0
[3sin²(θ/2) - 4sin(θ/2)cos(θ/2) + cos²(θ/2)] / cos²(θ/2)= 0
3tan²(θ/2) - 4tan(θ/2) + 1 = 0
[3tan(θ/2) - 1] [tan(θ/2) - 1] = 0
tan(θ/2) = 1/3 或 tan(θ/2) = 1

tanθ = 2tan(θ/2) / [1 - tan²(θ/2)]

當 tan(θ/2) = 1/3：
tanθ = 2(1/3) / [1 - (1/3)²] = 3/4

當 tan(θ/2) = 1：
tanθ = 2(1) / [1 - (1)²] = 2/0 (undefined)

所以 tanθ = 3/4

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