Question

若√(tan^2α-sin^2α)=tanα·sinα,则α的取值范围是？

Answer 1

√(tan^2α-sin^2α)=√[(sin^2α-sin^2αcos^2α)/cos^2α]=√(tan^2αsin^2α)=tanα·sinα
tanα·sinα>0，α为一、四象限角。即α属于(2kπ,2kπ+π/2)U(2kπ-π/2,2kπ)

Answer 2

tan^2α-sin^2α
=tan^2α-tan^2αcos^2α
=tan^2α(1-cos^2α)
=tan^2αsin^2α
√(tan^2α-sin^2α)=|tanα·sinα|=tanα·sinα
tanα·sinα>=0
α的取值范围是:
2kπ-π/2<α<2kπ+π/2或α=(2k+1)π. k∈Z

Answer 3

用a代替
tan²a-sin²a
=sin²a/cos²a-sin²a
=sin²a(1/cos²a-1)
=sin²a(1-cos²a)/cos²a
=sin²a*sin²a/cos²a
=sin²atan²a
即左边=√sin²atan²a=|sinatana|
右边=sinatana
所以sinatana≥0
sina*sina/cosa=sin²a/cosa≥0
则cosa>0
所以2kπ-π/2<a<2kπ+π/2

Answer 4

√(tan^2α-sin^2α)
＝√[sin^2a(1/cos^2a-1)]
=√[sin^2a(sec^2a-1)]
=√[sin^2a*tan^2a]
=tanα·sinα
说明tana和sina同号
kπ+π/2≤a≤kπ+π