0
$\begingroup$

$ \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?

  • 3
    Because $\sec(\theta) = \frac{1}{\cos(\theta)}$ by [definition](http://en.wikipedia.org/wiki/Trigonometric_functions#Reciprocal_functions).2011-11-18

1 Answers 1

2

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