eq b $ implies $ heta
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- S = 2i \cdot rac{e^{i( heta - \phi)/2} + e^{-i( heta - \phi)/2}}{e^{i( heta - \phi)/2} - e^{-i( heta - \phi)/2}} = 2i \cdot rac{2\cos\left( rac{ heta - \phi}{2}
- ight)}{2i\sin\left( rac{ heta - \phi}{2}
- ight)} = -2 \cot\left( rac{ heta - \phi}{2}
- eq \phi $, so $ S $ is purely imaginary. However, the original question may expect a real value. Re-examining:
- Let $ S = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a^2 - b^2} $. Let $ a = 1 $, $ b = i $:
- S = rac{2(1 + (-1))}{1 - (-1)} = 0, ext{ but } a^2 - b^2 = 1 - (-1) = 2, ext{ numerator } 2(1 + (-1)) = 0 \Rightarrow S = 0.
