Question

sinθ+cosθ=m,且tanθ+cotθ=n,则m,n的关系是

不好意思 忘记给大家选项了
A m^2=n
B m^2=2/n
c m^2=2/n+1
D m^2=2/n-1

Answer 1

解：根据：sin²θ+cos²θ=1
将 sinθ+cosθ=m 两边平方得则中：1+2sinθ*cosθ=m²----------------------(1)
又孙灶山因为，tanθ=sinθ/cosθ cotθ=cosθ/sinθ
所以， tanθ+cotθ=sinθ/cosθ + cosθ/sinθ=1/sinθ*cosθ =n
所以， sinθ*cosθ =1/n--------------------------(2)
由（1）和（2）辩链得：1+2/n=m²
故，选 (C)

Answer 2

tanθ+cotθ=1/陆汪(sinθcosθ) =n
根正明据举悉告sin2θ+cos2θ=1
(sinθ+cosθ)2－2sinθcosθ=1
m^2-2n=1

Answer 3

tanθ+cotθ = 1/凯简颂盯郑(sinθcosθ) =n
sinθcosθ = 1/n
(sinθ+cosθ)^2-2sinθcosθ = 1
故 m^2 = 2/n+1 选咐配 C

Answer 4

可用公式：棚告cotθ=cosθ/sinθ
tanθ=sinθ/樱和睁cosθ
tanθ+cotθ=cosθ/sinθ +sinθ/cosθ
=cosθ²+sinθ²/cosθ×sinθ
=1/m²-1
所以n=1/m²-1
m²n-n=1
m²脊岁=n+1/n

Answer 5

sin+cos=√2(sin+π/4)=m
tan+cot=sin/cos+cos/sin=(sin+cos)/厅敏sincos
sin+cos>=2√(sincos)
tan+cot>=2/(√sincos)
√sincos<1
tan+cot>2
所衡伏升以n>咐老m