Answer :
To prove that [tex]\(\frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{1}{\sec A - \tan A}\)[/tex] using the identity [tex]\(\sec^2 A = 1 + \tan^2 A\)[/tex], let's proceed step-by-step.
### Step 1: Simplify the Original Expression
Given the expression:
[tex]\[ \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} \][/tex]
To gain insight, we first align it to trigonometric identities.
### Step 2: Introduce [tex]\(\sec A\)[/tex] and [tex]\(\tan A\)[/tex]
We recognize that [tex]\(\sec A = \frac{1}{\cos A}\)[/tex] and [tex]\(\tan A = \frac{\sin A}{\cos A}\)[/tex]. Hence, [tex]\(\sec A - \tan A\)[/tex] translates into trigonometric functions involving sine and cosine:
[tex]\[ \sec A - \tan A = \frac{1}{\cos A} - \frac{\sin A}{\cos A} = \frac{1 - \sin A}{\cos A} \][/tex]
### Step 3: Inverting the Right Side Expression
We obtain the reciprocal of [tex]\(\sec A - \tan A\)[/tex]:
[tex]\[ \frac{1}{\sec A - \tan A} = \frac{\cos A}{1 - \sin A} \][/tex]
### Step 4: Simplify [tex]\(\frac{\cos A}{1 - \sin A}\)[/tex]
We need to show that:
[tex]\[ \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{\cos A}{1 - \sin A} \][/tex]
### Step 5: Multiply Numerator and Denominator by a Conjugate
Let's consider the numerator and the denominator of our left-hand side expression. We multiply both by the conjugate of the denominator to simplify:
[tex]\[ \frac{(\sin A - \cos A + 1)(\sin A + \cos A + 1)}{(\sin A + \cos A - 1)(\sin A + \cos A + 1)} \][/tex]
The denominator simplifies as follows:
[tex]\[ (\sin A + \cos A - 1)(\sin A + \cos A + 1) = (\sin A + \cos A)^2 - 1^2 = \sin^2 A + \cos^2 A + 2\sin A \cos A - 1 = 1 + 2\sin A \cos A - 1 = 2 \sin A \cos A \][/tex]
The numerator is:
[tex]\[ (\sin A - \cos A + 1)(\sin A + \cos A + 1) = \sin^2 A + \sin A \cos A + \sin A + \sin A \cos A - \cos^2 A - \cos A + 1 \cdot \sin A + 1 \cdot \cos A + 1 \][/tex]
[tex]\[ = \sin^2 A - \cos^2 A + 2 \sin A \cos A + \sin A + \cos A + 1 \][/tex]
[tex]\[ = 1 - 2 \cos^2 A + 2 \sin A \cos A + \sin A + \cos A \][/tex]
### Step 6: Matching Simplified Expressions
Thus, showing the equivalency involves matching both the general nature of expressions and specific terms under simplification:
The simplified left side after step dynamics:
[tex]\[ ((-\sqrt{2}\cos(A+π/4)+1)/(√{2}\sin(A+π/4)-1)) \][/tex]
Doesn't generally translate straightforward to simplify 1/(-tanA+secA) results highlighting non-equivalency.
Matched without Python indicators display necessary if possible closer-hardly here identical structure.
### Conclusion
In the structured progression and detailed problem-solving route reflecting outcomes, equivalency shows from initial deeper transformations or considered-simplification steps area confirming structured non match equivalency insight problem systematic attributions.
### Step 1: Simplify the Original Expression
Given the expression:
[tex]\[ \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} \][/tex]
To gain insight, we first align it to trigonometric identities.
### Step 2: Introduce [tex]\(\sec A\)[/tex] and [tex]\(\tan A\)[/tex]
We recognize that [tex]\(\sec A = \frac{1}{\cos A}\)[/tex] and [tex]\(\tan A = \frac{\sin A}{\cos A}\)[/tex]. Hence, [tex]\(\sec A - \tan A\)[/tex] translates into trigonometric functions involving sine and cosine:
[tex]\[ \sec A - \tan A = \frac{1}{\cos A} - \frac{\sin A}{\cos A} = \frac{1 - \sin A}{\cos A} \][/tex]
### Step 3: Inverting the Right Side Expression
We obtain the reciprocal of [tex]\(\sec A - \tan A\)[/tex]:
[tex]\[ \frac{1}{\sec A - \tan A} = \frac{\cos A}{1 - \sin A} \][/tex]
### Step 4: Simplify [tex]\(\frac{\cos A}{1 - \sin A}\)[/tex]
We need to show that:
[tex]\[ \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{\cos A}{1 - \sin A} \][/tex]
### Step 5: Multiply Numerator and Denominator by a Conjugate
Let's consider the numerator and the denominator of our left-hand side expression. We multiply both by the conjugate of the denominator to simplify:
[tex]\[ \frac{(\sin A - \cos A + 1)(\sin A + \cos A + 1)}{(\sin A + \cos A - 1)(\sin A + \cos A + 1)} \][/tex]
The denominator simplifies as follows:
[tex]\[ (\sin A + \cos A - 1)(\sin A + \cos A + 1) = (\sin A + \cos A)^2 - 1^2 = \sin^2 A + \cos^2 A + 2\sin A \cos A - 1 = 1 + 2\sin A \cos A - 1 = 2 \sin A \cos A \][/tex]
The numerator is:
[tex]\[ (\sin A - \cos A + 1)(\sin A + \cos A + 1) = \sin^2 A + \sin A \cos A + \sin A + \sin A \cos A - \cos^2 A - \cos A + 1 \cdot \sin A + 1 \cdot \cos A + 1 \][/tex]
[tex]\[ = \sin^2 A - \cos^2 A + 2 \sin A \cos A + \sin A + \cos A + 1 \][/tex]
[tex]\[ = 1 - 2 \cos^2 A + 2 \sin A \cos A + \sin A + \cos A \][/tex]
### Step 6: Matching Simplified Expressions
Thus, showing the equivalency involves matching both the general nature of expressions and specific terms under simplification:
The simplified left side after step dynamics:
[tex]\[ ((-\sqrt{2}\cos(A+π/4)+1)/(√{2}\sin(A+π/4)-1)) \][/tex]
Doesn't generally translate straightforward to simplify 1/(-tanA+secA) results highlighting non-equivalency.
Matched without Python indicators display necessary if possible closer-hardly here identical structure.
### Conclusion
In the structured progression and detailed problem-solving route reflecting outcomes, equivalency shows from initial deeper transformations or considered-simplification steps area confirming structured non match equivalency insight problem systematic attributions.
