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Prove that (Sinθ - Cosθ + 1)/(Sinθ + Cosθ - 1) = 1/(Secθ - Tanθ)

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Question

Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`

Sum
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Solution

LHS = `(sinθ - cosθ + 1)/(sinθ + cosθ - 1)`

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)`

LHS = `(sin^2θ + 1 + 2sinθ - cos^2θ)/(sin^2θ + cos^2θ + 2sinθcosθ - 1)`
 
LHS = `(1 - cos^2θ + 1 + 2sinθ - cos^2θ)/(1 + 2sinθcosθ - 1)   ...(sin^2θ + cos^2θ = 1)`
 
LHS = `(2 - 2cos^2θ + 2sinθ)/( 2sinθcosθ)`
 
LHS = `[cancel2(1 - cos^2θ + sinθ)]/[cancel2(sinθcosθ)]`
 
LHS = `( 1 - cos^2θ + sinθ)/(sinθcosθ)`
 
LHS = `(sin^2θ + sinθ)/( sinθcosθ )`
 
LHS = `(sinθ + 1)/cosθ`
 
LHS = `1/cosθ + sinθ/cosθ`
 
LHS = secθ + tanθ
 
LHS = `( secθ + tanθ ) xx (secθ - tanθ)/(secθ - tanθ)`
 
LHS = `(sec^2θ - tan^2θ)/(secθ - tanθ)`
 
LHS =`1/(secθ- tanθ)      ...[∴ sec^2θ − tan^2θ = 1]`
 
RHS = `1/(secθ- tanθ)`
 
LHS = RHS
 
Hence, it proved.
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Chapter 6: Trigonometry - Problem Set 6 [Page 138]

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