Trig question (1 Viewer)

[Wohzazz]

How can you convert

NB. alpha=@ beta=#

cos (@-#/2)sec(@+#/2)
= (1+ tan@/2. tan #/2)/(1-tan@/2.tan#/2)

 

[wogboy]

(1+ tan@/2. tan #/2)/(1-tan@/2.tan#/2)

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) }

 

[Wohzazz]

lol, i copied
cos(@-#/2).sec(@-#/2) instead of cos(@/2-#/2).sec(@/2-#/2)
anyways you got it again wogboy

'i'm with stupid'