Trig / Pre Cal/ identities Guide me please.
\( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x} + 1 \)

- anonymous

Trig / Pre Cal/ identities Guide me please.
\( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x} + 1 \)

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- anonymous

It is suppose to be.
\( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)

- anonymous

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

- anonymous

These things are killing me lol

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## More answers

- anonymous

not an identity

- xapproachesinfinity

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

- anonymous

I got that they're reciprocals

- xapproachesinfinity

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

- xapproachesinfinity

try some vlues if you have doubt

- anonymous

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

- anonymous

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

- xapproachesinfinity

now it does not work!
left does not equal right

- xapproachesinfinity

plug in some values to be sure if you have doubt

- xapproachesinfinity

no* for (now)

- anonymous

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

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

in that case it is an identity

- xapproachesinfinity

see that you give us the wrong thing from the start

- xapproachesinfinity

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

- anonymous

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?

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

you are making it hard

- anonymous

There are no rules though

- xapproachesinfinity

what rules?

- xapproachesinfinity

did you get what i did?

- anonymous

No

- xapproachesinfinity

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
\[=\frac{\sec x+1}{\tan x}\]

- xapproachesinfinity

perhaps i'm going fast?

- xapproachesinfinity

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

- anonymous

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.

- anonymous

For instance, how did you do it your way?

- xapproachesinfinity

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

- xapproachesinfinity

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

- anonymous

Yes that is my way.

- anonymous

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

- xapproachesinfinity

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

- anonymous

Correct

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

it is still working since top becomes sinx

- xapproachesinfinity

gotta go!
i think you got this now?

- anonymous

Ok so we have
\(\Huge \frac{(cosx)(tan x)}{cos x(\frac{1}{cos x}-1) }=\frac{\sec x+1}{\tan x } \)
\(\Huge \frac{sin x}{1-cos x}=\frac{\sec x+1}{\tan x } \)

- xapproachesinfinity

yes!

- xapproachesinfinity

i explained above how you obtain the right side

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