if m=costheta-sintheta and n=costheta+sintheta.show that √m/√n+√n√m=2/root over 1-tan²theta
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m=cosθ-sinθ and n=cosθ+sinθ
∴, m+ncosθ-sinθ+cosθ+sinθ=2cosθ and
mn=(cosθ-sinθ)(cosθ+sinθ)=cos²θ-sin²θ
LHS
=√m/√n+√n/√m
={(√m)²+(√n)²}/√(mn)
=(m+n)/√(mn)
=2cosθ/√(cos²θ-sin²θ)
=2/[{√(cos²θ-sin²θ)}/cosθ]
=2/√{(cos²θ-sin²θ)/cos²θ}
=2/√{1-(sin²θ/cos²θ)}
=2/√(1-tan²θ)
=RHS(Proved)
∴, m+ncosθ-sinθ+cosθ+sinθ=2cosθ and
mn=(cosθ-sinθ)(cosθ+sinθ)=cos²θ-sin²θ
LHS
=√m/√n+√n/√m
={(√m)²+(√n)²}/√(mn)
=(m+n)/√(mn)
=2cosθ/√(cos²θ-sin²θ)
=2/[{√(cos²θ-sin²θ)}/cosθ]
=2/√{(cos²θ-sin²θ)/cos²θ}
=2/√{1-(sin²θ/cos²θ)}
=2/√(1-tan²θ)
=RHS(Proved)
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