eq b $ implies $ heta

February 22, 2026

eq b $ implies $ heta

Related Articles

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.

Trending Articles