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Question
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Solution
LHS = `((sinθ - cosθ + 1)/(sinθ + cosθ - 1)) xx ((sinθ + cosθ + 1)/(sinθ + cosθ + 1))`
LHS = `((sinθ + 1 - cosθ )/(sinθ + cosθ - 1)) xx ((sinθ + 1 + cosθ)/(sinθ + cosθ + 1))`
LHS = `((sinθ + 1)^2 - cos^2θ)/((sinθ + cosθ)^2 - 1^2)`
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
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