Question

Trig / Pre Cal/ identities Guide me please.

\( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x} + 1 \)

Answer 1

It is suppose to be.

\( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)

Answer 2

\( \large \rm \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)

Answer 3

These things are killing me lol

Answer 4

not an identity

Answer 5

okay so we have \[\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\sec x+1}{\tan x }\]

Answer 6

I got that they're reciprocals

Answer 7

if so
we can do this \[\Huge \frac{\tan x}{\sec x+1}=\frac{\sec x+1}{\tan x}\]
clearly this is not truee

Answer 8

try some vlues if you have doubt

Answer 9

Yes but I am trying to get from the left side to the right.

\(\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\sec x+1}{\tan x }\)
Do I times by cos??

\(\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1} \times cos x\)

Answer 10

The identity is false is what he's saying. You won't be able to get from the left to the right.

Answer 11

now it does not work!
left does not equal right

Answer 12

plug in some values to be sure if you have doubt

Answer 13

no* for (now)

Answer 14

The original problem is
\(\Huge \frac{tan x}{\sec - 1 }=\frac{\sec x+1}{\tan x }\)

Answer 15

0 for example
is not good left give you 0
right gives you undefined 2/0
if it is an identity it should not be that way

Answer 16

I did the left side assuming that is what we have to do.

Answer 17

is the left
1/cosx -1 or 1/cosx +1
you just changed in your last reply?

Answer 18

in that case it is an identity

Answer 19

see that you give us the wrong thing from the start

Answer 20

if it is a minus on the bottom it is an identity
since tan^2x=sec^2x-1

Answer 21

This is the original problem.

\(\Huge \frac{tan x}{\sec - 1 }=\frac{\sec x+1}{\tan x } \)

How do we know to work the numerator or the denominator or both?

Answer 22

For instance, should I

\(\Huge \frac{tan x}{\frac{1}{cos x}-1 }=\frac{\sec x+1}{\tan x } \)

OR

\(\Huge \frac{\frac{sin}{cos}}{{\sec x}-1 }=\frac{\sec x+1}{\tan x } \)

OR
\(\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1} \)

Answer 23

usually when you prove something you should get from left to right of vice versa
so taking left
\[\frac{\tan x}{\sec x=1}=\frac{\tan x(\sec x+1)}{(\sec x-1)(\sec x+1)}\]
\[=\frac{\tan x(\sec x+1)}{\tan^2x }=\frac{\sec x+1}{\tan x} ~~~Q.E.D\]

Answer 24

you didn't need to do all that
just leave tan and sec that way

Answer 25

you are making it hard

Answer 26

There are no rules though

Answer 27

what rules?

Answer 28

did you get what i did?

Answer 29

No

Answer 30

if we do it your way we have
\[\frac{\sin x}{1-\cos x}=\frac{\sin x(1+\cos x)}{1-\cos^2x}=\frac{1+\cos x}{\sin x}\]
multiplyinh by 1/cos top and bottom we get
\[=\frac{\sec x+1}{\tan x}\]

Answer 31

perhaps i'm going fast?

Answer 32

i skip some stuff given that you know how to deal with algebra

Answer 33

I don't get it. How do I know to work the numerator or the denominator or both. How do I know what method to use when there are no rules to guy you.

Answer 34

For instance, how did you do it your way?

Answer 35

like \[\huge \frac{\frac{a}{b}}{\frac{1}{b}-1}=\frac{a}{b(\frac{1}{b}-1)}=\frac{a}{1-b}\]

Answer 36

this is what i did with your way changing tan to sin/cos and sec to 1/cos

Answer 37

Yes that is my way.

Answer 38

But there are other ways and some ways don't work

Answer 39

yes once you get sin x/1-cosx
you multiply by 1+cosx top and bottom

Answer 40

Correct

Answer 41

what other Ways do you mean
if you are using identities they should work

Answer 42

my way is no different than yours
i still had to multiply top and bottom be 1+sec x
to get sec^2x-1 on the bottom which is the same as tan^2x

Answer 43

this is just a matter of knowing your identities
and how to use them

Answer 44

For instance

\( \Huge \frac{tan x}{\frac{1}{cos x}-1 }=\frac{\sec x+1}{\tan x } \)

\( \Huge \frac{(cosx)(tan x)}{cos x(\frac{1}{cos x}-1) }=\frac{\sec x+1}{\tan x } \)

Answer 45

it is still working since top becomes sinx

Answer 46

gotta go!
i think you got this now?

Answer 47

Ok so we have

\(\Huge \frac{(cosx)(tan x)}{cos x(\frac{1}{cos x}-1) }=\frac{\sec x+1}{\tan x } \)

\(\Huge \frac{sin x}{1-cos x}=\frac{\sec x+1}{\tan x } \)

Answer 48

yes!

Answer 49

i explained above how you obtain the right side