Question

topic: PROVING IDENTITIES - help me please (SEE COMMENTS)
(trigonometry)

Answer 1

@Loser66 help

Answer 2

post your question please.

Answer 3

( Loser66 is offline )

Answer 4

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

Answer 5

i noticed

Answer 6

\[\huge\color{blue}{ \frac{1}{\sec(x)-\tan(x)} =\sec(x)+\tan(x) } \]

Answer 7

\[\huge\color{blue}{ 1 =(\sec(x)+\tan(x) )(\sec(x)-\tan(x))} \]

Answer 8

I multiplied both sides times sec(40)-tan(40)

Answer 9

why? is that possible?

i thought we can only manipulate 1 side?

Answer 10

shouldn't the other side be stable?

Answer 11

Oh, then I take what I wrote back.

Answer 12

okay

Answer 13

\[\huge\color{blue}{ \frac{1}{\sec(x)-\tan(x)}=\sec(x)+\tan(x) } \]
LEFT SIDE:\

top and bottom times

sec(x)+tan(x)

\[\huge\color{blue}{ \frac{\sec(x)-\tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) } \]

Now, use the identity for the bottom,
\[\sec^2x=\tan^2x+1\]

Answer 14

Oh, the top should say + not - sorry.

Answer 15

\[\huge\color{blue}{ \frac{\sec(x) \huge\color{red}{ + } \tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) } \]

Answer 16

HUGE HINT: \[\sec^2x=\tan^2x+1~~~~~~~ther efore~~~~~\sec^2x-\tan^2x=1\]

Answer 17

i do not get it. sorry

Answer 18

Do you get what I did before.

the last huge post in blue with a red plus ?

Answer 19

nope

but hey i found this:
http://symbolab.com/solver/trigonometric-identity-calculator/prove%20%5Cfrac%7B1%7D%7B%5Csec(x)-%5Ctan(x)%7D%3D%5Csec(x)%2B%5Ctan(x)

but the answer doesnt seem to be equal
is there a way you could make it equal???

Answer 20

Don't try to look it up else where, follow me.
lets start over, ok?
\[\huge\color{green} { \frac{1}{\sec(x)-\tan(x)}=\sec(x)+\tan(x) }\]

I am working the left side:

1) Multiply top and bottom times sec(x)+tan(x)

Answer 21

ok

Answer 22

\[\color{green} { \frac{\sec(x)+\tan(x)}{(~~\sec(x)-\tan(x)~~)~(~~\sec(x)+\tan(x)~~)}=\sec(x)+\tan(x) }\]
\[\color{green} { \frac{\sec(x)+\tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) }\]

tell me if you don't get it.

Answer 23

okay

Answer 24

Now lets use an identity.

\[\tan^2(x)+1=\sec^2(x)\]
let's draw some conclusions.
\[\tan^2(x)+1\color{blue} { -\tan^2(x) } =\sec^2x\color{blue} { -\tan^2(x) }\]\[1=\sec^2x-\tan^2x\]

look at the denominator, what is it now equal to?
has your identity been verified?

Answer 25

yes. got it

thanks @SolomonZelman

Answer 26

You welcome, no problem, just knowing identities that's all the skill involved.

Answer 27

and wow. that was shorter

Answer 28

Yeah, I explained the same thing, but in a shorter way.

Answer 29

what's the technique?

Answer 30

how come you're so good at this?

Answer 31

what do you mean, the technique for what?

Ohhh, I took trig last year, and got 97. I am just good at math and bad at other staff.

Answer 32

technique in answering / proving these trigonometric identities?

you did it so fast and in a shorter and much efficient way

that's really cool

WOW! Congratz! good for you

Answer 33

Thank you! I am sure that if you try to do a lot of practice problems you will master them.

Answer 34

i really hope so. failing trigonometry is the last thing that i'd want

Answer 35

WHAT?! FAILING?! what is your current grade may I ask ?

Answer 36

last semester i got 2.0 in algebra

now i'm taking trigonometry and it do is very hard. i dont have my grade yet since the 2nd semester just started.

Answer 37

1 being the highest
3 - pass
5 - fail

Answer 38

@SolomonZelman ^