#sin^2(x)+cos^2(x) = 1#
divide everything by #cos(x)# ...we'll worry about the case when #color(red)(cos(x) = 0)# later
#color(white)("XXXX")##color(blue)(sin^2(x)/cos(x) + cos(x) = 1/cos(x))#
since #tan(x) = sin(x)/cos(x)# and # sec(x) = 1/cos(x)#
#color(white)("XXXX")##color(blue)(sin(x)tan(x) + cos(x) = sec(x))#
rearrange to look closer to our target equation:
#color(white)("XXXX")##color(blue)(cos(x) + tan(x)sin(x) = sec(x))#
looks like we need to add #sin(x)# to make the left side look like the target
#color(white)("XXXX")##color(blue)(sin(x) +cos(x)+tan(x)sin(x) = sec(x) + sin(x))#
since #tan(x) = sin(x)/cos(x)#
therefore #sin(x) = cos(x)tan(x)
#color(white)("XXXX")##color(blue)(sin(x)+cos(x)+tan(x)sin(x) = sec(x) + cos(x)tan(x))#
We have now proven the target equation except for the case #color(red)(cos(x) = 0)#
If #cos(x) = 0# then #tan(x)# is undefined
and since #tan(x)# appears on both sides
#color(white)("XXXX")#we will pretend that #"undefined" = "undefined"#
#color(white)("XXXX")#is adequate to cover this case.
