化简根号下[(1-cos x)/(1+cos x)] + 根号下[(1+cos x)/(1-cos x)]其中x属于(π/2,π)

问题描述:

化简根号下[(1-cos x)/(1+cos x)] + 根号下[(1+cos x)/(1-cos x)]其中x属于(π/2,π)

根号下[(1-cos x)/(1+cos x)] + 根号下[(1+cos x)/(1-cos x)]
=(1-cos x)/sin x+(1+cos x)/sin x=2/sin x

1-cos x=2(sin2分之x )^2 1+cos x=2(cos2分之x )^2 x属于(π/2,π) sin2分之x >0 cos2分之x

√(1-cosx)(1+cosx)+√(1+cosx)(1-cosx)
=2√(1-cosx)(1+cosx)
=2√(1-cosx^2)
=2√(Sinx^2)
x∈(π/2,π)
=2Sinx