If m=(cosθ−sinθ) and n=(cosθ+sinθ) then show that
√mn+√nm=2√1−tan2θ
√mn+√nm=√(cosθ−sinθ)(cosθ+sinθ+√(cosθ+sinθ(cosθ−sinθ)
=√(cosθ−sinθ)(cosθ+sinθ×√(cosθ−sinθ(cosθ−sinθ)+√(cosθ+sinθ(cosθ−sinθ)×√(cosθ+sinθ)(cosθ+sinθ
=(cosθ−sinθ)√cos2θ−sin2θ+cosθ+sinθ√cos2θ−sin2θ
=(2cosθ)√cos2θ−sin2θ
=2√cos2θ√cos2θ−sin2θ
=2√1−tan2θ