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

Sum

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, 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
Is there an error in this question or solution?

#### APPEARS IN

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