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$$ \Rightarrow (\sec^2 \theta) \frac{d \theta}{dt} = \frac{A \frac{dB}{dt} - B \frac{dA}{dt}}{A^2} $$ $$ \Rightarrow \frac{d \theta}{dt} = (\cos^2 \theta) \left(\frac{A \frac{dB}{dt} - B \frac{dA}{dt}}{A^2} \right) $$

I have this. One question. I think I should divide by $\sec^2 \theta$. But I saw this solution to the problem. Why does sec become cos?

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Because $\sec(\theta) = \frac{1}{\cos(\theta)}$ by definition. – Sasha Nov 18 '11 at 16:42

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By definition $\sec \theta = \frac{1}{\cos \theta}$, so it disappears from the left-hand side when you multiply by the cosine.

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