#### Question

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

#### 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, sin

^{2}θ + cos^{2}θ = 1Therefore,

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

`= (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,

sec

Therefore,

=`1/(secθ- tanθ)`

Hence, it proved.

sec

^{2}θ - tan^{2}θ = 1Therefore,

=`1/(secθ- tanθ)`

Hence, it proved.

Is there an error in this question or solution?

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Prove that (Sinθ - Cosθ + 1)/(Sinθ + Cosθ - 1) = 1/(Secθ - Tanθ) Concept: Application of Trigonometry.

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