Question

已知3π/2<α<2π,化简：{[√(1-cosα)]+[√(1+cosα)]}/{[√(1-cosα)-√(1+cosα)]}

已知3π/2<α<2π,化简：{[√(1-cosα)]+[√(1+cosα)]}/{[√(1-cosα)-√(1+cosα)]}+[√(1+sinα)]/[√(1-sinα)].

Answer 1

：{[√(1-cosα)]+[√(1+cosα)]}/{[√(1-cosα)-√(1+cosα)]}+[√(1+sinα)]/[√(1-sinα)].
因为:{[√(1-cosα)]+[√(1+cosα)]}/{[√(1-cosα)-√(1+cosα)]}
={[√(1-cosα)]+[√(1+cosα)]}/{[√(1-cosα)-√(1+cosα)]}*[√(1-cosα)]+[√(1+cosα)]}/{[√(1-cosα)+√(1+cosα)]}
=={[√(1-cosα)]+[√(1+cosα)]}*[√(1-cosα)]+[√(1+cosα)]}/{根号(1-cos^2α)}
={(1-cosα)]+(1+cosα)+2√(1-cosα)(1+cosα)}/(-sina)
=(2-2sina)/(-sina) (注意sina<0)

又[√(1+sinα)]/[√(1-sinα)].=(1+sina)/根号(1-sin^2a)=(1+sina)/cosa (cosa>0)
原式=(2sina-2)/sina+(1+sina)/cosa
=2-2/sina+tga+1/cosa