Let's prove the equation step-by-step:
[tex]\[ \frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \sec A + \csc A \][/tex]
1. Let [tex]\( x = \cos A + \sin A \)[/tex]:
The equation can be rewritten as:
[tex]\[ \frac{1}{x - 1} + \frac{1}{x + 1} = \sec A + \csc A \][/tex]
2. Combine the left-hand side using a common denominator:
The common denominator of the fractions on the left-hand side is [tex]\((x - 1)(x + 1)\)[/tex]. With that common denominator, let's combine the terms:
[tex]\[ \frac{1}{x - 1} + \frac{1}{x + 1} = \frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)} \][/tex]
3. Simplify the numerator:
[tex]\[ \frac{x + 1 + x - 1}{(x^2 - 1)} = \frac{2x}{x^2 - 1} \][/tex]
4. Express [tex]\( x^2 \)[/tex] in terms of trigonometric functions:
Since [tex]\( x = \cos A + \sin A \)[/tex], let's calculate [tex]\( x^2 \)[/tex]:
[tex]\[ x^2 = (\cos A + \sin A)^2 \][/tex]
[tex]\[ x^2 = \cos^2 A + 2 \cos A \sin A + \sin^2 A \][/tex]
[tex]\[ x^2 = (\cos^2 A + \sin^2 A) + 2 \cos A \sin A \][/tex]
[tex]\[ x^2 = 1 + 2 \cos A \sin A \][/tex]
5. Substitute [tex]\( x^2 \)[/tex] back into the denominator of our expression:
[tex]\[ \frac{2(\cos A + \sin A)}{1 + 2 \cos A \sin A - 1} = \frac{2(\cos A + \sin A)}{2 \cos A \sin A} \][/tex]
6. Simplify the fraction:
[tex]\[ \frac{2(\cos A + \sin A)}{2 \cos A \sin A} = \frac{\cos A + \sin A}{\cos A \sin A} \][/tex]
7. Separate the right-hand side into two terms:
[tex]\[ \frac{\cos A + \sin A}{\cos A \sin A} = \frac{\cos A}{\cos A \sin A} + \frac{\sin A}{\cos A \sin A} \][/tex]
[tex]\[ = \frac{1}{\sin A} + \frac{1}{\cos A} \][/tex]
[tex]\[ = \csc A + \sec A \][/tex]
Thus, we've proven that:
[tex]\[ \frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \sec A + \csc A \][/tex]
This completes the proof.
