Expression eqivalent to 1+sec/cos ? Explain

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Homework Help Overview

The discussion revolves around finding an expression equivalent to \( \frac{1+\sec\theta}{\cos\theta} \) and simplifying it to \( \frac{\cos\theta+1}{\cos^2\theta} \). The subject area includes trigonometric identities and simplification techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various methods for simplifying the expression, including substituting \( \sec\theta \) with \( \frac{1}{\cos\theta} \) and finding common denominators. Some participants express confusion about the simplification process and seek clarification on specific steps.

Discussion Status

There is an ongoing exploration of different approaches to simplify the expression. Some participants have provided guidance on using properties of fractions and suggested rewriting the expression in various forms. However, there is no explicit consensus on a single method, and participants continue to question and clarify their understanding.

Contextual Notes

Some participants mention having difficulty with similar types of problems in the past, indicating a potential lack of familiarity with trigonometric identities and simplification techniques. There are also references to previous discussions and advice given in related threads.

aisha
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Expression eqivalent to 1+sec/cos ?? Explain please

How do u find an expression equivalent to [tex]\frac {1+\sec\theta} {\cos\theta}[/tex] to be [tex]\frac {\cos\theta+1} {\cos^2\theta}[/tex]

When I did this question I got my answer to be [tex]\frac {2} {\cos\theta}[/tex]

Im not very good with identities can someone please explain to me how to do this question and other like it.
 
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Can you show your steps to getting that? I'll point out where you went wrong.
 
[tex]\frac {sin^2\theta + \cos^2\theta} {\cos \theta} + \frac {\sin^2\theta+\cos^2\theta} {\cos\theta} (this is over \cos \theta)[/tex]
 
aisha said:
[tex]\frac {sin^2\theta + \cos^2\theta} {\cos \theta} + \frac {\sin^2\theta+\cos^2\theta} {\cos\theta} (this is over \cos \theta)[/tex]

Forget about [tex]sin^2\theta + \cos^2\theta =1[/tex].

Do the problem again, but this time only use:[tex]sec\theta=\frac{1}{cos\theta}[/tex] and try to simplify it.
 
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I have gone through so many of these types of questions but I have no idea of how to simplify it, I've looked everywhere for a similar example but just can't find anything, I don't know what to do... :cry: How do u solve these questions that ask for equivalent identities? :redface:
 
Well, did you take learningphysics' advice? Substitute 1/sec(x) every time you see cos(x) to get the expression in terms of secant, instead of cosine. That is definitely a step in the right direction.

BTW, check back to your thread "please explain". I posted the solution to your previous problem there, since you were very close to understanding it anyway.
 
Use:
[tex]\sec x =\frac{1}{\cos x}[/tex]

and
[tex]\frac{1+\frac{1}{a}}{b}=\frac{a+1}{ab}[/tex]

which is an elementary property of fractions.

Daniel.
 
dextercioby said:
Use:
[tex]\sec x =\frac{1}{\cos x}[/tex]

and
[tex]\frac{1+\frac{1}{a}}{b}=\frac{a+1}{ab}[/tex]

which is an elementary property of fractions.

Daniel.

Can u explain how u got this elementary property of fractions, this might be the step I don't remember lol
 
aisha said:
I have gone through so many of these types of questions but I have no idea of how to simplify it, I've looked everywhere for a similar example but just can't find anything, I don't know what to do... :cry: How do u solve these questions that ask for equivalent identities? :redface:

I plugged in [tex]sec\theta=\frac{1}{cos\theta}[/tex]

I get:

[tex]\frac {1+\frac{1}{cos\theta}} {\cos\theta}[/tex]

Do something to the numerator and denominator so that you get a cleaner looking expression. Try different things... you can try adding the two terms in the numerator together (get a common denominator etc...)
 
  • #10
learningphysics said:
I plugged in [tex]sec\theta=\frac{1}{cos\theta}[/tex]

I get:

[tex]\frac {1+\frac{1}{cos\theta}} {\cos\theta}[/tex]

Do something to the numerator and denominator so that you get a cleaner looking expression. Try different things... you can try adding the two terms in the numerator together (get a common denominator etc...)

This is exactly where I am stuck!I tried multiplying the numerator and denominator by cos theta, but that didnt work, I took ur advice and I am half way I made the numerators have a common denominator

so I got [tex]1+\frac{1}{\cos\theta} = \frac {\cos\theta} {\cos\theta}+ \frac {1}{\cos\theta}[/tex] This is only the numerator of the question my answer was [tex]\frac {\cos\theta+1} {\cos\theta}[/tex] But this is all over [tex]\cos\theta[/tex] what do u do next? How do you get cos^2theta in the denominator of the answer ?
 
  • #11
aisha said:
Can u explain how u got this elementary property of fractions, this might be the step I don't remember lol

Try working on the expression dextercioby gave... how can you rewrite the expression... try experimenting...

That 1/a is messy, how can you rewrite the expression so that the 1/a is no longer there?
 
  • #12
aisha said:
Can u explain how u got this elementary property of fractions, this might be the step I don't remember lol

[tex]\frac{1+\frac{1}{a}}{b}=\frac{\frac{a+1}{a}}{b}=\frac{a+1}{ab}[/tex]

Daniel.
 
  • #13
aisha said:
This is exactly where I am stuck!I tried multiplying the numerator and denominator by cos theta, but that didnt work, I took ur advice and I am half way I made the numerators have a common denominator

so I got [tex]1+\frac{1}{\cos\theta} = \frac {\cos\theta} {\cos\theta}+ \frac {1}{\cos\theta}[/tex] This is only the numerator of the question my answer was [tex]\frac {\cos\theta+1} {\cos\theta}[/tex] But this is all over [tex]\cos\theta[/tex] what do u do next? How do you get cos^2theta in the denominator of the answer ?

Ok. I see where you are stuck now. The whole fraction is:
[tex]\frac { \frac {\cos\theta} {\cos\theta}+ \frac {1}{\cos\theta}}{\cos\theta}[/tex]

I'll rewrite the numerator, and the fraction becomes:
[tex]\frac { \frac {\cos\theta +1}{\cos\theta}}{cos\theta}[/tex]

Now the denominator of this fraction is [tex]\cos\theta[/tex]

Instead of dividing by [tex]\cos\theta[/tex], I'm going to multiply by:
[tex]\frac{1}{\cos\theta}[/tex]

So we have:
[tex]\frac {\cos\theta +1}{\cos\theta} * \frac{1}{\cos\theta}[/tex]

Multiply the numerators, and multiply the denominators and we get:
[tex]\frac {\cos\theta+1}{\cos^2\theta}[/tex]
 
  • #14
[tex]\frac{(\frac{a}{b})}{c} = \frac{a}{bc}[/tex]
 
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  • #15
Why not just multiple top and bottom by cos...
[tex]\frac {1+\sec\theta} {\cos\theta}[/tex]
[tex]\frac {1+\sec\theta} {\cos\theta}(\frac{\cos\theta}{\cos\theta})[/tex]
[tex]\frac {\cos\theta+1} {\cos^2\theta}[/tex]

because cos*sec =1 = cos*(1/cos) = 1
 
  • #16
:eek: WoW I think I get it thanks sooo much everyone, once again u all saved me!
 

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