To evaluate the given expression [tex]\(\frac{\sec A - \tan A}{\sec A + \tan A}\)[/tex] and compare it against [tex]\(1 - 2 \sec A \cdot \tan A + 2 \tan^2 A\)[/tex], let's break it down step by step:
1. Start with the original expression:
[tex]\[
\frac{\sec A - \tan A}{\sec A + \tan A}
\][/tex]
2. Simplify by multiplying the numerator and the denominator by the conjugate of the denominator:
This will involve multiplying the numerator and the denominator by [tex]\((\sec A - \tan A)\)[/tex]:
[tex]\[
\frac{(\sec A - \tan A)(\sec A - \tan A)}{(\sec A + \tan A)(\sec A - \tan A)}
\][/tex]
3. Apply the difference of squares:
The denominator simplifies using the difference of squares:
[tex]\[
(\sec A + \tan A)(\sec A - \tan A) = \sec^2 A - \tan^2 A
\][/tex]
For the numerator, we use:
[tex]\[
(\sec A - \tan A)^2 = \sec^2 A - 2 \sec A \tan A + \tan^2 A
\][/tex]
So the expression becomes:
[tex]\[
\frac{\sec^2 A - 2 \sec A \tan A + \tan^2 A}{\sec^2 A - \tan^2 A}
\][/tex]
4. Recall the Pythagorean identity:
[tex]\[
\sec^2 A = 1 + \tan^2 A
\][/tex]
Substituting this into our denominator:
[tex]\[
\sec^2 A - \tan^2 A = (1 + \tan^2 A) - \tan^2 A = 1
\][/tex]
So our expression simplifies to:
[tex]\[
\frac{\sec^2 A - 2 \sec A \tan A + \tan^2 A}{1} = \sec^2 A - 2 \sec A \tan A + \tan^2 A
\][/tex]
5. Substitute [tex]\(\sec^2 A = 1 + \tan^2 A\)[/tex] back in the simplified numerator:
[tex]\[
\sec^2 A - 2 \sec A \tan A + \tan^2 A = (1 + \tan^2 A) - 2 \sec A \tan A + \tan^2 A
\][/tex]
Simplifying further gives:
[tex]\[
1 + \tan^2 A + \tan^2 A - 2 \sec A \tan A = 1 + 2 \tan^2 A - 2 \sec A \tan A
\][/tex]
6. Final simplified form:
We end up with the simplified expression:
[tex]\[
1 - 2 \sec A \tan A + 2 \tan^2 A
\][/tex]
Therefore, although we initially set out to prove that:
[tex]\[
\frac{\sec A - \tan A}{\sec A + \tan A} \stackrel{?}{=} 1 - 2 \sec A \tan A + 2 \tan^2 A
\][/tex]
It turns out that it simplifies to:
[tex]\[
\frac{\sec A - \tan A}{\sec A + \tan A} = 1 - 2 \sec A \tan A + 2 \tan^2 A \text{ only if } \sec(A)=\cos(A) \text{ and } \tan(A)=\sin(A)
\][/tex]
Thus, their equality is generally false unless a specific case occurs.
