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Question

If sin A + cos A = m and

sec A + cosec A = n, show that :

n(m21)=2m


Solution

sinθ+cosθ=m,secθ+cosecθ=n
Consider LHS
n(m21)=(secθ+cosecθ)[(cosθ+sinθ)21]=(secθ+cosecθ)[cos2θ+sin2θ+2sinθcosθ1]=(secθ+cosecθ)[2sinθcosθ]since cos2θ+sin2θ=1=secθ.2sinθcosθ+cosecθ.2sinθcosθ=2sinθ+2cosθ=2[sinθ+cosθ]=2m

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