Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

nbedwards934

  • 2 years ago

tanθ+cotθ=cscθxsecθ (I will draw what I have so far)

  • This Question is Closed
  1. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1354253316580:dw|

  2. jidery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1354253520763:dw|

  3. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Are you trying to prove this identity??

  4. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes I am. I've started working on the side cscθxsecθ but I'm stuck. I drew as far as I got in the first drawing.

  5. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Would you like further help?

  6. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes please

  7. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Very well. Let me start from the beginning them and go step by step.

  8. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    alright. Thank you

  9. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so. i'll use x instead of theta. start by replacing tanx with (sin/cos) and cotx with (cos/sin) so... (cosx/sinx)+(sinx/cosx)=cscxsecx now replace cscx with (1/sinx) and secx with (1/cos) (cosx/sinx)+(sinx/cosx)=(1/cosx)(1/sinx)

  10. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Are you following so far?

  11. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you're working with both sides? My professor told us to only work with one side and leave the other as is.

  12. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Alright, If you want, I can work only on one side.

  13. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that would be better, thank you. I started work on the cscθxsecθ side

  14. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    okay, would you mind if I work from the left side?

  15. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I actually might have figured this one out. Give me just a minute to finish it

  16. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok

  17. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    tanx+cotx=cscxsecx Replace tanx with an equivalent expression (sinx)/(cosx) using the fundamental identities. (sinx)/(cosx)+cotx=cscxsecx Replace (cosx)/(sinx) using the fundamental identities. (sinx)/(cosx)+(cosx)/(sinx)=cscxsecx To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is cosxsinx. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. (sinx)/(cosx)*(sinx)/(sinx)+(cosx)/(sinx)*(cosx)/(cosx)=cscxsecx Multiply sinx by sinx to get sinx^(2). (sin^(2)x)/(sinx*cosx)+(cosx)/(sinx)*(cosx)/(cosx)=cscxsecx Multiply cosx by sinx to get sinxcosx. (sin^(2)x)/(sinxcosx)+(cosx)/(sinx)*(cosx)/(cosx)=cscxsecx Multiply cosx by cosx to get cosx^(2). (sin^(2)x)/(sinxcosx)+(cos^(2)x)/(cosx*sinx)=cscxsecx Multiply sinx by cosx to get sinxcosx. (sin^(2)x)/(sinxcosx)+(cos^(2)x)/(sinxcosx)=cscxsecx Combine the numerators of all expressions that have common denominators. (sin^(2)x+cos^(2)x)/(sinxcosx)=cscxsecx Replace sin^(2)x+cos^(2)x with 1 using the identity sin^(2)(x)+cos^(2)(x)=1. (1)/(sinxcosx)=cscxsecx Replace the expressions with an equivalent expression using the fundamental identities. (1)/(sinxcosx)....Replace with cscxsecx Therefore cscxsecx=cscxsecx

  18. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Does that help?

  19. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'll be honest, it's very difficult to interpret the equations you have typed up there, but I'm trying haha

  20. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Here. This might help a bit. It's solved by a computer.

  21. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

  22. nbedwards934
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so much! I didn't even think about changing the sin^2θ+cos^2θ to 1. I have the formula right in front of me, but it didn't register. But now it makes sense. Thanks!

  23. zordoloom
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep

  24. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.