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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is suppose to be. \( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\( \large \rm \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0These things are killing me lol

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2okay so we have \[\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\sec x+1}{\tan x }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got that they're reciprocals

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2if so we can do this \[\Huge \frac{\tan x}{\sec x+1}=\frac{\sec x+1}{\tan x}\] clearly this is not truee

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2try some vlues if you have doubt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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
 one year ago
Best ResponseYou've already chosen the best response.0The identity is false is what he's saying. You won't be able to get from the left to the right.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2now it does not work! left does not equal right

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2plug in some values to be sure if you have doubt

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2no* for (now)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The original problem is \(\Huge \frac{tan x}{\sec  1 }=\frac{\sec x+1}{\tan x }\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.20 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
 one year ago
Best ResponseYou've already chosen the best response.0I did the left side assuming that is what we have to do.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2is the left 1/cosx 1 or 1/cosx +1 you just changed in your last reply?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2in that case it is an identity

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2see that you give us the wrong thing from the start

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2if it is a minus on the bottom it is an identity since tan^2x=sec^2x1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0For 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
 one year ago
Best ResponseYou've already chosen the best response.2usually when you prove something you should get from left to right of vice versa so taking left \[\frac{\tan x}{\sec x=1}=\frac{\tan x(\sec x+1)}{(\sec x1)(\sec x+1)}\] \[=\frac{\tan x(\sec x+1)}{\tan^2x }=\frac{\sec x+1}{\tan x} ~~~Q.E.D\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2you didn't need to do all that just leave tan and sec that way

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2you are making it hard

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There are no rules though

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2what rules?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2did you get what i did?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.2perhaps i'm going fast?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2i skip some stuff given that you know how to deal with algebra

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0For instance, how did you do it your way?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2like \[\huge \frac{\frac{a}{b}}{\frac{1}{b}1}=\frac{a}{b(\frac{1}{b}1)}=\frac{a}{1b}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2this is what i did with your way changing tan to sin/cos and sec to 1/cos

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But there are other ways and some ways don't work

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2yes once you get sin x/1cosx you multiply by 1+cosx top and bottom

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2what other Ways do you mean if you are using identities they should work

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2my way is no different than yours i still had to multiply top and bottom be 1+sec x to get sec^2x1 on the bottom which is the same as tan^2x

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2this is just a matter of knowing your identities and how to use them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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
 one year ago
Best ResponseYou've already chosen the best response.2it is still working since top becomes sinx

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2gotta go! i think you got this now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so we have \(\Huge \frac{(cosx)(tan x)}{cos x(\frac{1}{cos x}1) }=\frac{\sec x+1}{\tan x } \) \(\Huge \frac{sin x}{1cos x}=\frac{\sec x+1}{\tan x } \)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2i explained above how you obtain the right side
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