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Home » Since \(x^2 + y^2 + z^2 = 1\), we have: - Midis

Since \(x^2 + y^2 + z^2 = 1\), we have: - Midis

📅 March 12, 2026 👤 scraface

Mar 12, 2026

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x^2 - 2x + y^2 + z^2 = 0 \implies (x^2 - 2x + 1) + y^2 + z^2 = 1 \impl...
This is a sphere centered at $(1, 0, 0)$ with radius 1.
\boxed{\text{A sphere of radius 1 centered at } (1, 0, 0)}
Question: Find the maximum value of $\left(\sin x + \csc x\right)^2 +...
Solution: Expand the expression:
Simplify using $\sin^2 x + \cos^2 x = 1$ and $\csc^2 x = 1 + \cot^2 x$...
+ 2 + 2 + \cot^2 x + \tan^2 x = 5 + \cot^2 x + \tan^2 x.
Let $t = \tan^2 x$, so $\cot^2 x = \frac{1}{t}$. The expression become...
+ t + \frac{1}{t}.
By AM-GM, $t + \frac{1}{t} \geq 2$, with equality when $t = 1$ (i.e.,...

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