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Set equal to -1: - Londonproperty

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 Note that:

📌 a^2 + b^2 = b^2(x^2 + 1), \quad a^2 - b^2 = b^2(x^2 - 1)

📌 So the expression becomes:

📌 \frac{b^2(x^2 + 1)}{b^2(x^2 - 1)} + \frac{b^2(x^2 - 1)}{b^2(x^2 + 1)} = \frac{x^2 + 1}{x^2 - 1} + \frac{x^2 - 1}{x^2 + 1}

📌 Let $ y = \frac{x^2 + 1}{x^2 - 1} $, then the expression becomes:

📌 y + \frac{1}{y}

📌 We compute $ y + \frac{1}{y} $. First, recall that $ x + \frac{1}{x} = 5 $. Square both sides:

📌 x^2 + 2 + \frac{1}{x^2} = 25 \Rightarrow x^2 + \frac{1}{x^2} = 23

🔥 Popular Posts

- 2\sin^2\theta = \sin \theta
Rearrange into a quadratic in \(\sin \theta\):
2\sin^2\theta + \sin \theta - 1 = 0
Let \( x = \sin \theta \), so:
2x^2 + x - 1 = 0
x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{1 +...
This gives:
x = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-4}{4} =...
So, \( \sin \theta = \frac{1}{2} \) or \( \sin \theta = -1 \).
Solve for \( \theta \in [0^\circ, 360^\circ] \):

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