If sin θ+cos θ=m and sec θ+cosec θ=n, prove that
n(m2−1)=2m.
Answer:
- Substituting sin θ+cos θ=m and sec θ+cosec θ=n in n(m2−1), we have n(m2−1)=sec θ+cosec θ ((sin θ+cos θ)2−1)=(sec θ+cosec θ)(sin2 θ+cos2 θ+2sin θcos θ−1)=(1cos θ+1sin θ)(1+2sin θcos θ−1)[∵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
- Hence, n(m2−1) = 2m.