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sec 240 degrees
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sec 240=sec 270-30
its -2 but i don't know what to use to get there?
right but its 1/cos..
or wait nvm...im gonna try it
1/cos(270-30) =1/cos(3*90 -30)
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cos 270-theta=-sin theta
is this true: sec(u-v)=sec u sec v + csc u csc v
so here it will be nothing but \[\dfrac{-1}{\sin 30 }\]
why should u think of that when u hav straight method available?
:)
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i know but im trying to make sense of it in my head
@ghazi
@sauravshakya
1/cos(270-30) =1/cos(3*90 -30)=1/(-sin30)
oh wait yes ^ that makes more sense
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make a reading of this for future clarity http://www.sparknotes.com/math/trigonometry/trigonometricfunctions/section4.rhtml
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