QQuestion
Show that \text{tan θ + sec θ - 1}\over \text{tan θ - sec θ + 1} = 1 + \text{sin θ}\over \text{cos θ}
✓Answer
LHS: \text{tan θ + sec θ - 1}\over \text{tan θ - sec θ + 1}
= \text{(tan θ + sec θ) - (sec² θ - tan² θ)}\over \text{tan θ - sec θ + 1}
= \text{(sec θ + tan θ) - (sec θ + tan θ)(sec θ - tan θ)}\over \text{tan θ - sec θ + 1}
= \text{(sec θ + tan θ)(1 - sec θ + tan θ)}\over \text{tan θ - sec θ + 1}
= sec θ + tan θ
= 1\over \text{cos θ} + \text{sin θ}\over \text{cos θ}
= 1 + \text{sin θ}\over \text{cos θ}
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