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

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Sum

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

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Solution

Consider the L.H.S.

`(sinθ - cosθ + 1)/(sinθ + cosθ - 1)`

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

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

= `((sinθ + 1)^2 - cos^2θ)/((sinθ + cosθ)^2 - 1^2)`

= `(sin^2θ + 1 + 2sinθ - cos^2θ)/(sin^2θ + cos^θ + 2sinθcosθ - 1)`
since, sin2θ + cos2θ = 1
Therefore,
`= (1 - cos^2θ + 1 + 2sinθ - cos^2θ)/(1 + 2sinθcosθ - 1)`
 
`= (2 - 2cos^2θ + 2sinθ)/( 2sinθcosθ)`
 
`=( 1 - cos^2θ + sinθ)/(sinθcosθ)`
 
`=(sin^2θ + sinθ)/( sinθcosθ )`
 
`=(sinθ + 1)/cosθ`
 
`= 1/cosθ + sinθ/cosθ`
 
= secθ + tanθ
 
= `( secθ + tanθ ) xx (secθ - tanθ)/(secθ - tanθ)`
 
= `(sec^2θ - tan^2θ)/(secθ - tanθ)`
We know that,
sec2θ - tan2θ = 1
Therefore,
=`1/(secθ- tanθ)`
Hence, it proved.
Concept: Application of Trigonometry
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APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 6 Trigonometry
Problem Set 6 | Q 5.1 | Page 138
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