Prove that (sinA - cosA + 1) / (sinA + cosA - 1) = 1 / (secA - tanA)

We need to prove:

(sin A - cos A + 1) / (sin A + cos A - 1) = 1 / (sec A - tan A)

Step-by-step proof:

Simplify the expression for sec A and tan A:

sec A = 1 / cos A
tan A = sin A / cos A
Thus, sec A - tan A = (1 / cos A) - (sin A / cos A) = (1 - sin A) / cos A.

Rewrite the RHS in terms of sine and cosine:

The RHS is: 1 / (sec A - tan A)
= 1 / [(1 - sin A) / cos A]
= cos A / (1 - sin A).

Simplify the LHS:

LHS = (sin A - cos A + 1) / (sin A + cos A - 1).

Multiply numerator and denominator by the conjugate of the denominator, (sin A + cos A - 1):

Numerator: (sin A - cos A + 1)(sin A + cos A - 1)
= (sin A)^2 - (cos A)^2 + sin A - cos A + sin A + 1
= sin²A - cos²A + 2sin A - cos A + 1.

Denominator: (sin A + cos A - 1)²
= sin²A + cos²A - 2sin A - 2cos A + 1
= 1 - 2sin A - 2cos A + 1
= 2(1 - sin A - cos A).

So, LHS = (sin²A - cos²A + 2sin A - cos A + 1) / [2(1 - sin A - cos A)].

Factorize numerator if possible:
We observe symmetry at completion ratios