What is \(\tan(1125°)\cot(405°) + \tan(765°)\cot(675°)\) equal to?
0
Reducing each angle modulo 360°: \(1125° \equiv 45°\), \(405° \equiv 45°\), \(765° \equiv 45°\) and \(675° \equiv 315°\). So \(\tan(1125°)\cot(405°) = \tan 45°\cot 45° = 1\) and \(\tan(765°)\cot(675°) = \tan 45°\cot 315° = 1 \times (-1) = -1\). Adding gives \(1 + (-1) = 0\).
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