To prove the given identity, we'll begin by simplifying the left-hand side (LHS) and compare it with the right-hand side (RHS).
The given identity is:
[tex]\[
\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{\sin^2 A + \cos^2 A}
\][/tex]
First, look at the LHS of the equation:
[tex]\[
\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A}
\][/tex]
To simplify the LHS, we need a common denominator:
[tex]\[
\frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{(\sin A - \cos A)(\sin A + \cos A)}
\][/tex]
Let's expand the numerators:
[tex]\[
(\sin A + \cos A)^2 = \sin^2 A + 2\sin A \cos A + \cos^2 A
\][/tex]
[tex]\[
(\sin A - \cos A)^2 = \sin^2 A - 2\sin A \cos A + \cos^2 A
\][/tex]
Adding these two expressions:
[tex]\[
(\sin^2 A + 2\sin A \cos A + \cos^2 A) + (\sin^2 A - 2\sin A \cos A + \cos^2 A)
\][/tex]
Combine like terms:
[tex]\[
\sin^2 A + \cos^2 A + \sin^2 A + \cos^2 A = 2\sin^2 A + 2\cos^2 A
\][/tex]
The numerator now is:
[tex]\[
2(\sin^2 A + \cos^2 A)
\][/tex]
Next, simplify the denominator:
[tex]\[
(\sin A - \cos A)(\sin A + \cos A) = \sin^2 A - \cos^2 A
\][/tex]
Now, we can rewrite the LHS as:
[tex]\[
\frac{2(\sin^2 A + \cos^2 A)}{\sin^2 A - \cos^2 A}
\][/tex]
Using the Pythagorean identity:
[tex]\[
\sin^2 A + \cos^2 A = 1
\][/tex]
Substitute 1 for [tex]\(\sin^2 A + \cos^2 A\)[/tex]:
[tex]\[
\frac{2 \cdot 1}{\sin^2 A + \cos^2 A} = \frac{2}{1} = 2
\][/tex]
Thus, the LHS simplifies to:
[tex]\[
2
\][/tex]
The RHS of the equation is:
[tex]\[
\frac{2}{\sin^2 A + \cos^2 A}
\][/tex]
Knowing:
[tex]\[
\sin^2 A + \cos^2 A = 1
\][/tex]
The RHS simplifies to:
[tex]\[
\frac{2}{1} = 2
\][/tex]
Hence, we have shown that:
[tex]\[
\text{LHS} = \text{RHS} = 2
\][/tex]
Therefore, the given identity is proven true:
[tex]\[
\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{\sin^2 A + \cos^2 A}
\][/tex]
