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Home » But no option gives -1. - jntua results

But no option gives -1. - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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But earlier steps suggest $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. For $...
S = rac{2(e^{2i heta} + e^{2i\phi})}{e^{2i heta} - e^{2i\phi}} = 2 \c...
Multiply numerator and denominator by $ e^{-i( heta + \phi)/2} $:
S = 2i \cdot rac{e^{i( heta - \phi)/2} + e^{-i( heta - \phi)/2}}{e^{i...
ight)}{2i\sin\left(rac{ heta - \phi}{2}
ight)} = -2 \cot\left(rac{ heta - \phi}{2}
eq b $ implies $ heta
eq \phi $, so $ S $ is purely imaginary. However, the original questio...
Let $ S = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a...
S = rac{2(1 + (-1))}{1 - (-1)} = 0, ext{ but } a^2 - b^2 = 1 - (-1)...

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