We are trying to prove that #(cos t)/(1+sint) + (1+sin t)/(cos t) = 2 sect#. We're going to need these trig identities:
# cos^2x+sin^2x=1 #
# secx=1/cosx #
Now here's the proof. I'll start with the left side and manipulate it until it looks exactly like the right side. First, get a common denominator between the fractions, combine them, and then simplify:
# LHS=(cos t)/(1+sint) + (1+sin t)/(cos t) #
# color(white)(LHS)=(costcolor(red)(*cost))/((1+sint)color(red)(\*cost))+((1+sint)color(red)(\*(1+sint)))/(costcolor(red)(\*(1+sint))) #
# color(white)(LHS)=(cos^2t)/(cost(1+sint))+(1+2sint+sin^2t)/(cost(1+sint)) #
# color(white)(LHS)=(cos^2t+1+2sint+sin^2t)/(cost(1+sint)) #
# color(white)(LHS)=(color(red)cancelcolor(black)(cos^2t+sin^2t)^1+1+2sint)/(cost(1+sint)) #
# color(white)(LHS)=(2+2sint)/(cost(1+sint)) #
# color(white)(LHS)=(2(1+sint))/(cost(1+sint)) #
# color(white)(LHS)=(2color(red)cancelcolor(black)((1+sint)))/(costcolor(red)cancelcolor(black)((1+sint))) #
# color(white)(LHS)=2/cost #
# color(white)(LHS)=2*1/cost #
# color(white)(LHS)=2*sect #
# color(white)(LHS)=2sect #
# color(white)(LHS)=RHS #
That's what we were trying to prove. Hope this helped!
