Answer
See the explanation below.
Work Step by Step
$\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$
And the identity has to be further simplified with the help of multiplying with the same fraction
$\begin{align}
& \frac{\sin x-\cos x+1}{\sin x+\cos x-1}=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}\times \frac{\sin x-\cos x-1}{\sin x-\cos x-1} \\
& =\frac{{{\sin }^{2}}-2\cos x\sin x+{{\cos }^{2}}x-1}{{{\sin }^{2}}x-2\sin x-{{\cos }^{2}}x+1} \\
& =\frac{{{\sin }^{2}}x+{{\cos }^{2}}x-2\cos x\sin x-1}{\sin {{x}^{2}}-2\sin x-\left( 1-{{\sin }^{2}}x \right)+1} \\
& =\frac{1-2\cos x\sin x-1}{{{\sin }^{2}}x-2\sin x+\sin {{x}^{2}}}
\end{align}$
Further solving the expression:
$\begin{align}
& \frac{1-2\cos x\sin x-1}{{{\sin }^{2}}x-2\sin x+\sin {{x}^{2}}}=\frac{-2\cos x\sin x}{2{{\sin }^{2}}x-2\sin x} \\
& =\frac{-2\sin x\cos x}{2\sin x\left( \sin x-1 \right)} \\
& =\frac{-\cos x}{\sin x-\cos x} \\
& =\frac{-\cos x}{\sin x-1}\times \frac{\sin x+1}{\sin x+1}
\end{align}$
Further simplification of expression:
$\begin{align}
& \frac{-\cos x}{\sin x-1}\times \frac{\sin x+1}{\sin x+1}=\frac{-\cos x\left( \sin x+1 \right)}{{{\sin }^{2}}x-1} \\
& =\frac{-\cos x\left( \sin x+1 \right)}{-{{\cos }^{2}}x} \\
& =\frac{-\cos x\left( \sin x+1 \right)}{-{{\cos }^{2}}x} \\
& =\frac{\sin x+1}{\cos x}
\end{align}$
Hence, the left side of the expression is equal to the right side, which is $\frac{\sin x-\cos x+1}{\sin x+\cos x-1}=\frac{\sin x+1}{\cos x}$.
