Since $ |a| = |b| = 1 $, $ a \overline{a} = 1 $ and $ b \overline{b} = 1 $. However, observe that $ a^2 - b^2 = (a - b)(a + b) $, but no simplification arises directly. Instead, compute $ S + \overline{S} $:

February 22, 2026

$Since $ |a| = |b| = 1 $, $ a \overline{a} = 1 $ and $ b \overline{b} = 1 $. However, observe that $ a^2 - b^2 = (a - b)(a + b) $, but no simplification arises directly. Instead, compute $ S + \overline{S} $:$

eq b $. Find the value of $ rac{a + b}{a - b} + rac{a - b}{a + b} $.
Solution: Let $ S = rac{a + b}{a - b} + rac{a - b}{a + b} $. Combine the fractions:
S = rac{(a + b)^2 + (a - b)^2}{(a - b)(a + b)} = rac{2a^2 + 2b^2}{a^2 - b^2}.
S + \overline{S} = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a^2 - b^2}.
But $ a
eq b $, and $ |a| = |b| = 1 $. Let $ b = \overline{a} $ (since $ |a| = 1 $), but $ a

Since $ |a| = |b| = 1 $, $ a \overline{a} = 1 $ and $ b \overline{b} = 1 $. However, observe that $ a^2 - b^2 = (a - b)(a + b) $, but no simplification arises directly. Instead, compute $ S + \overline{S} $:

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