anonymous
  • anonymous
Trig / Pre Cal/ identities Guide me please. \( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x} + 1 \)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
It is suppose to be. \( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)
anonymous
  • anonymous
\( \large \rm \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)
anonymous
  • anonymous
These things are killing me lol

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anonymous
  • anonymous
not an identity
xapproachesinfinity
  • xapproachesinfinity
okay so we have \[\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\sec x+1}{\tan x }\]
anonymous
  • anonymous
I got that they're reciprocals
xapproachesinfinity
  • 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
  • xapproachesinfinity
try some vlues if you have doubt
anonymous
  • 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
  • anonymous
The identity is false is what he's saying. You won't be able to get from the left to the right.
xapproachesinfinity
  • xapproachesinfinity
now it does not work! left does not equal right
xapproachesinfinity
  • xapproachesinfinity
plug in some values to be sure if you have doubt
xapproachesinfinity
  • xapproachesinfinity
no* for (now)
anonymous
  • anonymous
The original problem is \(\Huge \frac{tan x}{\sec - 1 }=\frac{\sec x+1}{\tan x }\)
xapproachesinfinity
  • 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
  • anonymous
I did the left side assuming that is what we have to do.
xapproachesinfinity
  • xapproachesinfinity
is the left 1/cosx -1 or 1/cosx +1 you just changed in your last reply?
xapproachesinfinity
  • xapproachesinfinity
in that case it is an identity
xapproachesinfinity
  • xapproachesinfinity
see that you give us the wrong thing from the start
xapproachesinfinity
  • xapproachesinfinity
if it is a minus on the bottom it is an identity since tan^2x=sec^2x-1
anonymous
  • 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
  • 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
  • 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
  • xapproachesinfinity
you didn't need to do all that just leave tan and sec that way
xapproachesinfinity
  • xapproachesinfinity
you are making it hard
anonymous
  • anonymous
There are no rules though
xapproachesinfinity
  • xapproachesinfinity
what rules?
xapproachesinfinity
  • xapproachesinfinity
did you get what i did?
anonymous
  • anonymous
No
xapproachesinfinity
  • 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
  • xapproachesinfinity
perhaps i'm going fast?
xapproachesinfinity
  • xapproachesinfinity
i skip some stuff given that you know how to deal with algebra
anonymous
  • 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
  • anonymous
For instance, how did you do it your way?
xapproachesinfinity
  • xapproachesinfinity
like \[\huge \frac{\frac{a}{b}}{\frac{1}{b}-1}=\frac{a}{b(\frac{1}{b}-1)}=\frac{a}{1-b}\]
xapproachesinfinity
  • xapproachesinfinity
this is what i did with your way changing tan to sin/cos and sec to 1/cos
anonymous
  • anonymous
Yes that is my way.
anonymous
  • anonymous
But there are other ways and some ways don't work
xapproachesinfinity
  • xapproachesinfinity
yes once you get sin x/1-cosx you multiply by 1+cosx top and bottom
anonymous
  • anonymous
Correct
xapproachesinfinity
  • xapproachesinfinity
what other Ways do you mean if you are using identities they should work
xapproachesinfinity
  • 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
  • xapproachesinfinity
this is just a matter of knowing your identities and how to use them
anonymous
  • 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
  • xapproachesinfinity
it is still working since top becomes sinx
xapproachesinfinity
  • xapproachesinfinity
gotta go! i think you got this now?
anonymous
  • 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
  • xapproachesinfinity
yes!
xapproachesinfinity
  • xapproachesinfinity
i explained above how you obtain the right side

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