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Solution:

Correct Answer:

$\frac{\tanθ + \secθ - 1}{\tanθ - \secθ + 1}$

$\frac{\frac{\sinθ}{\cosθ} + \frac{1}{\cosθ} - 1}{\frac{\sinθ}{\cosθ} - \frac{1}{\cosθ} + 1}$

$\frac{\sinθ - \cosθ + 1}{\sinθ + \cosθ - 1}$

$\frac{\sinθ - \cosθ + 1}{\sinθ + \cosθ - 1}$ $\times$ $\frac{\sinθ + \cosθ + 1}{\sinθ + \cosθ + 1}$

$\frac{1 + 2\sinθ + {\sin}^2θ - {\cos}^2θ}{ {\sin}^2θ + {\cos}^2θ+ 2\sinθ \cosθ - 1}$

$\frac{2\sinθ + 2{\sin}^2θ}{2\sinθ \cosθ}$

$\frac{2\sinθ (1 + \sinθ) }{2\sinθ \cosθ}$

$\frac{1 + \sinθ}{\cosθ}$

$\frac{\tanθ + \secθ - 1}{\tanθ - \secθ + 1}$

$\frac{\frac{\sinθ}{\cosθ} + \frac{1}{\cosθ} - 1}{\frac{\sinθ}{\cosθ} - \frac{1}{\cosθ} + 1}$

$\frac{\sinθ - \cosθ + 1}{\sinθ + \cosθ - 1}$

$\frac{\sinθ - \cosθ + 1}{\sinθ + \cosθ - 1}$ $\times$ $\frac{\sinθ + \cosθ + 1}{\sinθ + \cosθ + 1}$

$\frac{1 + 2\sinθ + {\sinθ}^2 - {\cosθ}^2}{ {\sinθ}^2 + {\cosθ}^2 + 2\sinθ \cosθ - 1}$

$\frac{2\sinθ + 2{\sinθ}^2 } { 2\sinθ \cosθ }$

$\frac{2\sinθ (1 + \sinθ) }{ 2\sinθ \cosθ }$

$\frac{1 + \sinθ}{\cosθ}$