1
Question
Prove that
sec θ+tan θ−1tan θ−sec θ+1=cos θ(1−sin θ) [4 MARKS]
sec θ+tan θ−1tan θ−sec θ+1=cos θ(1−sin θ) [4 MARKS]
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Solution
Formula: 1 Mark
Concept: 1 Mark
Application: 2 Marks
sec θ+tan θ−1tan θ−sec θ+1
=(sec θ+tan θ)−(sec2 θ−tan2 θ)(tan θ−sec θ+1) [∵sec2 θ−tan2 θ=1]
=(sec θ+tan θ)[1−(sec θ−tan θ)](tan θ−sec θ+1)
=(sec θ+tan θ)(tan θ−sec θ+1)(tan θ−sec θ+1)=(sec θ+tan θ)
=(1cos θ+sin θcos θ)=(1+sin θ)cos θ=(1+sin θ)cos θ×(1−sin θ)(1−sin θ)
=(1−sin2 θ)cos θ(1−sin θ)=cos2 θcos θ(1−sin θ)=cos θ(1−sin θ)
Hence,
sec θ+tan θ−1tan θ−sec θ+1=cos θ(1−sin θ)
Concept: 1 Mark
Application: 2 Marks
sec θ+tan θ−1tan θ−sec θ+1
=(sec θ+tan θ)−(sec2 θ−tan2 θ)(tan θ−sec θ+1) [∵sec2 θ−tan2 θ=1]
=(sec θ+tan θ)[1−(sec θ−tan θ)](tan θ−sec θ+1)
=(sec θ+tan θ)(tan θ−sec θ+1)(tan θ−sec θ+1)=(sec θ+tan θ)
=(1cos θ+sin θcos θ)=(1+sin θ)cos θ=(1+sin θ)cos θ×(1−sin θ)(1−sin θ)
=(1−sin2 θ)cos θ(1−sin θ)=cos2 θcos θ(1−sin θ)=cos θ(1−sin θ)
Hence,
sec θ+tan θ−1tan θ−sec θ+1=cos θ(1−sin θ)
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