We will use the Identity # : sec^2A-tan^2A=1#.
Let us divide the #Nr.# and #Dr.# of L.H.S by #cosA# to get,
The L.H.S. #={(sinA+1-cosA)/cosA}/{(sinA-1+cosA)/cosA#
#={sinA/cosA+1/cosA-cosA/cosA}/{sinA/cosA-1/cosA+cosA/cosA}#
#=(tanA+secA-1)/(tanA-secA+1}#
#={tanA+secA-(sec^2A-tan^2A)}/(tanA-secA+1)#
#={(tanA+secA)-(secA+tanA)(secA-tanA)}/(tanA-secA+1)#
#={(secA+tanA)cancel((1-secA+tanA))}/cancel((tanA-secA+1)#
#=secA+tanA#
#=# The R.H.S.
Hence, the Proof. Enjoy Maths.!
#II^(nd)# Method :-
We have, #1-sin^2A=cos^2A#
#rArr (1+sinA)(1-sinA)=cos^2A#
#rArr (1+sinA)/cosA=cosA/(1-sinA)#
#:.# Each Rratio #={(1+sinA)-(cosA)}/{(cosA)-(1-sinA)}#, i.e.,
Each Ratio#=(1+sinA-cosA)/(cosA-1+sinA)#.
In particular, #(1+sinA)/cosA=(1+sinA-cosA)/(cosA-1+sinA)#, or,
#1/cosA+sinA/cosA=(1+sinA-cosA)/(cosA-1+sinA)#.
#rArr secA+tanA=(1+sinA-cosA)/(cosA-1+sinA)#.