sec x + tan x = 1 / (sec x - tan x)?

Question

another proof. i've tried everything i can think of, but to no effect. just a hint will be appreciated. thanks in advance

Answer 1

from your trig identites 1 + tan^2 x = sec^2 x

rearranging gives: sec^2 x - tan^2 x = 1

now sec^2 x - tan^2 x can be wrote like (sec x + tan x)(secx - tanx)

so we can now divide both sides by (secx - tanx)
which gives (secx + tanx) = 1/(secx - tanx)

if you have all your trig identities to begin with in front of you while you do these proofs, it is much easier to get the hang of them.

pick me as your best answer!

Answer 2

You probably know that,
cos^2 x + sin^2 x = 1
divide throughout by cos^2 x, then
1 + tan^2 x = sec^2 x
as sec^2 x = 1/cos^2 x

then sec^2 x - tan^2 x = 1
(sec x -tan x) (sec x + tan x) =1
divide by (sec x - tan x) and you get the desired result.

the main identity here is the Pythagorean theorem of trigonometry.
try,
http://en.wikipedia.org/wiki/Trigonometric_identity
http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity

Answer 3

sin^2 x + cos^2 x =1
then cos^2 x = 1 - sin^2 x = ( 1 - sin x )( 1 + sin x)
Divide both sides by cos x ( 1 - sin x ) and get
cos x / (1 - sin x ) = (1 + sin x )/ cos x = (1/cos x ) + (sin x / cos x)
= sec x + tan x
Divide the numerator and denominator of the left side both by cos x
(cos x/cos x)/ ((1 - sin x)/cos x) = 1/ ((1/cos x) - (sin x/cos x))
= 1/ (sec x - tan x)

Answer 4

sec x = 1/cosX
tan x = sinx / cosx

You can rewrite your equation as:
1/cox + sinx/cosx = 1/(1/cosx - sinx/cosx)
That simplifies to
(1+sinx)/cosx = 1/((1-sinx)/cosx)

Now get the reciprocal of the denominator of the last term and multiply:
(1+sinx)/cosx = 1 * (cosx/(1-sinx))
(1+sinx)/cosx = cosx/(1-sinx)

Cross multiply
cosx(1+sinx) = cosx(1-sinx)

Divide both sides by cosx

1+sinx = 1-sinx.

Wait, that's wrong I think, sorry. I might have made a mistake somewhere??

Answer 5

Left hand side
1/cosx+sinx/cosx
1=sinx/cosx
both up and down side multiply with (1-sinx)
1-sin^2 x/cosx(1-sinx)

cos^2 x/cosx(1-sinx)

cosx/(1-sinx)

both up and down side divide by cos X

1/Secx-tanX

*this is kind of okay question boss.u shld've tried.

Answer 6

There is definitely a "tan" in the answer. Good Luck! :)

Answer 7

sec x is 1:cos x, so "sec x + tan x = 1 / (sec x - tan x)" is the same as
"1/cos x + sin x /cos x =1/(1/cos x -sin x /cos x )"
And with basic transformations:
(1+sin x):cos x=1:((1-sin x):cos x)
(1+sin x):cos x=cos x:(1-sin x)
(1+sin x)(1-sin x)=cos x*cos x
1-sin x*sin x=cos x*cos x
1=sin x*sin x+cos x*cos x
...
And that is true

Answer 8

s*2-t*2=1
1/t*2-t*2=1
1-t*4=t*2
if t*2=p
p*2+p-1=0

Answer 9

sec^2x-tan^2x=1
1/cos^2x - sin^2x =1
REJOINED THEIR DENOMINATOR
1-sin^2x/cos^2=1
cos^2x/cos^2x=1 [1-sin^2x=cos^2x]
henced proved

Answer 10

UMMM, YEAH, IM NOT THAT SMART. GOOD LUCK!