text: stuck on trigonometric identities.? how can i prove that: 1/(sec x + tan x)  = sec x - tan x  what i tried to do (but i can't get to the prove) is:  1/ (1/cos + sin/cos)  = (1/1)/(1+sen/cos) = cos / 1 + sen = 1 / (sec + sen) = cos + csc  what procedure do you suggest so i can get to sec x - tan x ? thanks.  also, is there any way to check my procedures in mathematica for students? i remember having trouble with these types of problems too.  there's so many trig identities to choose from, it's hard to know sometimes what to use!  since there's no trig identity (that i know of) for secx + tanx, i think i'd multiply my fraction by (secx-tanx) / (secx-tanx) to see if i can get rid of the denominator.  (a trick that sometimes works when you have a-b in the denominator is to multiply top and bottom by a+b so that you get a²-b² in the denominator.):  1 / (secx + tanx) =   (secx - tanx) / (secx + tanx)(secx - tanx) =  (secx - tanx) / (sec²x - tan²x)  now i would use the trig identity tan²x + 1 = sec²2 (because notice this can be rewritten as 1 = sec²x - tan²x).  so ....  (secx - tanx) / (sec²x - tan²x) =   (secx - tanx) / 1 =  secx - tanx  hope that helped!  ~ ♥ ~