Stuck on trigonometric identities.?

Question

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?

Answer 1

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!

~ ♥ ~

Answer 2

1/(sec x + tan x) = sec x - tan x
1/[(1/cos x) + (sin x/cos x)]
(1/cos x) + (sin x/cos x)
sec x + tan x

umm.... i dont know how it equals sec x - tan x
sorry

Answer 3

multiply (sec x-tan x)/(sec x-tan x)
you will have (sec x)^2 -(tan x)^2=1 in your denominator, in your numerator you have sec x-tan x
so it is proved.

Answer 4

Well tan²X + 1 = sec²X

So sec²X - tan²X = 1
So (secX - tanX)(secX + tanX) = 1
Thus 1/(secX + tanX) = secX - tanX

Answer 5

1+cos²x=2-sin²x Add -1 to both side: cos²x = 1 - sin²x cos²x + sin²x = 1 checked sin²t - cos²t = 2 sin²t - 1 Add 2cos²t to both side: sin²t + cos²t = 2(sin²t + cos²t) - 1 = 2(1) -1 = 1 again checked