If sin θ+cos θ=m and sec θ+cosec θ=n, prove that n(m21)=2m.


Answer:


Step by Step Explanation:
  1. Substituting sin θ+cos θ=m and sec θ+cosec θ=n in n(m21), we have n(m21)=sec θ+cosec θ ((sin θ+cos θ)21)=(sec θ+cosec θ)(sin2 θ+cos2 θ+2sin θcos θ1)=(1cos θ+1sin θ)(1+2sin θcos θ1)[sin2 θ+cos2 θ=1,sec θ=1cos θ, and cosec θ=1sin θ]=(sin θ+cos θsin θcos θ)(2sin θcos θ)=2(sin θ+cos θ)=2m
  2. Hence, n(m21) = 2m.

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