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Parametics and conics... (1 Viewer)

Michaelmoo

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Ok. So I'm on question 5, exercise 3.3 from cambridge. Looking at the solutions, I can't see how they get this:

asec@ - acos@ = acos@tan^2@

Is there a rule/theorem I havn't touched on?

Thanks.
 

azureus88

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2009
not really, its just:

asec@ - acos@
=acos@(sec^2@ - 1)
=acos@tan^2@
 

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