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Home » "Miranda Cosgrove in a Banana Bikini: Shocking Beach Look She Never Revealed! - Groen Casting

"Miranda Cosgrove in a Banana Bikini: Shocking Beach Look She Never Revealed! - Groen Casting

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 S = rac{2(1 + (-1))}{1 - (-1)} = 0.

📌 But earlier steps suggest $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. For $ |a| = |b| = 1 $, $ a^2 \overline{a}^2 = 1 $, so $ a^2 = rac{1}{\overline{a}^2} $. This path is complex. Instead, let $ a = 1 $, $ b = -1 $: $ S = rac{0}{2} + rac{2}{0} $, invalid. Correct approach: Let $ a = e^{i heta} $, $ b = e^{i\phi} $, then

📌 S = rac{2(e^{2i heta} + e^{2i\phi})}{e^{2i heta} - e^{2i\phi}} = 2 \cdot rac{e^{i heta} + e^{i\phi}}{e^{i heta} - e^{i\phi}}.

📌 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( 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 b $ implies $ heta

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