Prove : (sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)

Question

Prove :

\(\frac{sin\theta - cos\theta + 1}{sin\,\theta + cos\,\theta - 1} = \frac{1}{sec\,\theta - tan\,\theta}\)

(sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)

Answer 1

We know that, 

sin²θ + cos²θ = 1 

∴ 1 – sin²θ = cos²θ 

∴ (1 – sin θ) (1 + sin θ) = cos θ. cos θ

∴ By theorem on equal ratios,

Consider,

From (i) and (ii), we get

Alternate method :

...[Dividing numerator and denominator by cos θ]

Answer 2

LHS = (sin θ - cos θ + 1)/(sin θ + cos θ - 1)

=cos θ (sin θ - cos θ + 1)/cos θ(sin θ + cos θ - 1)

=cos θ (sin θ - cos θ + 1)/cos θ[cos θ - (1-sin θ)]

=cos θ (sin θ - cos θ + 1)/[cos² θ -cos θ (1-sin θ)]

=cos θ (sin θ - cos θ + 1)/[(1-sin² θ) -cos θ (1-sin θ)]

=cos θ (sin θ - cos θ + 1)/[(1-sin θ)(1+sin θ)-cos θ (1-sin θ)]

=cos θ (sin θ - cos θ + 1)/(1-sin θ)[(1+sin θ)-cos θ ]

=cos θ (sin θ - cos θ + 1)/(1-sin θ)[(sin θ -cos θ +1)]

=cos θ /(1-sin θ)

=1 /[(1-sin θ)/cos θ]

=1 /[(1/cos θ-sin θ/cos θ]

= 1/(sec θ - tan θ)= RHS

proved