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Trig question (1 Viewer)
- Thread starter Wohzazz
- Start date
I think you mean (1 + tan@tan#/2)/(1 - tan@tan#/2) instead, don't you?
I'll use A and B instead of @ and # for less confusion.
cos(A - B/2) * sec(A + B/2)
= cos(A - B/2) / cos(A + B/2)
= { cos(A)*cos(B/2) + sin(A)*sin(B/2) } / { cos(A)*cos(B/2) - sin(A)*sin(B/2) }
divide top and bottom by cos(A)*cos(B/2):
= { 1 + tan(A)*tan(B/2) } / { 1 - tan(A)*tan(B/2) }