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Home » \(a = -2\), \(b = 40\) - Crosslake

\(a = -2\), \(b = 40\) - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Question: Find all angles $ z \in [0^\circ, 360^\circ] $ such that $ \tan z + \cot z = 2 $.

📌 Solution: Start with $ \tan z + \cot z = 2 $. Recall $ \tan z = \frac{\sin z}{\cos z} $, $ \cot z = \frac{\cos z}{\sin z} $. Then:

📌 \tan z + \cot z = \frac{\sin z}{\cos z} + \frac{\cos z}{\sin z} = \frac{\sin^2 z + \cos^2 z}{\sin z \cos z} = \frac{1}{\sin z \cos z}

📌 Set $ \frac{1}{\sin z \cos z} = 2 $. Then $ \sin z \cos z = \frac{1}{2} $. Using the identity $ \sin(2z) = 2\sin z \cos z $, we get $ \sin(2z) = 1 $. Solve $ \sin(2z) = 1 $:

📌 2z = 90^\circ + 360^\circ n \Rightarrow z = 45^\circ + 180^\circ n

📌 Within $ [0^\circ, 360^\circ] $, the solutions are $ z = 45^\circ $ and $ z = 225^\circ $. Thus, the angles are $ \boxed{45^\circ \text{ and } 225^\circ} $.

📌 Question: Find the minimum value of $ (\tan x + \cot x)^2 + (\sec x + \csc x)^2 $ for $ x \in (0^\circ, 90^\circ) $.

📌 Solution: Let’s simplify each term. First, $ \tan x + \cot x = \frac{\sin x}{\cos x} + \frac

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