Let's examine the given equation and verify whether it holds true:
[tex]\[
\frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \sec A + \csc A.
\][/tex]
### Step 1: Simplify the left-hand side (LHS) of the equation.
Let [tex]\( x = \cos A + \sin A \)[/tex]. Then we can rewrite the LHS as:
[tex]\[
\frac{1}{x - 1} + \frac{1}{x + 1}.
\][/tex]
### Step 2: Combine the fractions on the LHS.
We combine the fractions by obtaining a common denominator:
[tex]\[
\frac{1}{x - 1} + \frac{1}{x + 1} = \frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)} = \frac{x + 1 + x - 1}{(x - 1)(x + 1)} = \frac{2x}{x^2 - 1}.
\][/tex]
Recall that [tex]\( x = \cos A + \sin A \)[/tex], so substituting [tex]\( x \)[/tex] back into the fraction:
[tex]\[
\frac{2(\cos A + \sin A)}{(\cos A + \sin A)^2 - 1}.
\][/tex]
### Step 3: Simplify the expression.
[tex]\[
(\cos A + \sin A)^2 = \cos^2 A + \sin^2 A + 2 \cos A \sin A = 1 + 2 \cos A \sin A.
\][/tex]
Then the denominator becomes:
[tex]\[
(\cos A + \sin A)^2 - 1 = (1 + 2 \cos A \sin A) - 1 = 2 \cos A \sin A.
\][/tex]
Therefore, the LHS simplifies to:
[tex]\[
\frac{2(\cos A + \sin A)}{2 \cos A \sin A} = \frac{\cos A + \sin A}{\cos A \sin A}.
\][/tex]
We can split this fraction into two separate terms:
[tex]\[
\frac{\cos A + \sin A}{\cos A \sin A} = \frac{\cos A}{\cos A \sin A} + \frac{\sin A}{\cos A \sin A} = \frac{1}{\sin A} + \frac{1}{\cos A} = \csc A + \sec A.
\][/tex]
### Step 4: Compare the simplified LHS with the right-hand side (RHS).
So, we have:
[tex]\[
\frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \csc A + \sec A.
\][/tex]
Thus, we see that the LHS simplifies directly to the RHS:
[tex]\[
\sec A + \csc A = \sec A + \csc A.
\][/tex]
Hence, we have shown that the given equation holds true.
[tex]\[
\boxed{\frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \sec A + \csc A}.
\][/tex]
