## nbedwards934 2 years ago tanθ+cotθ=cscθxsecθ (I will draw what I have so far)

1. nbedwards934

|dw:1354253316580:dw|

2. jidery

|dw:1354253520763:dw|

3. zordoloom

Are you trying to prove this identity??

4. nbedwards934

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

Would you like further help?

6. nbedwards934

7. zordoloom

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

8. nbedwards934

alright. Thank you

9. zordoloom

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

Are you following so far?

11. nbedwards934

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

12. zordoloom

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

13. nbedwards934

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

14. zordoloom

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

15. nbedwards934

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

16. zordoloom

ok

17. zordoloom

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

Does that help?

19. nbedwards934

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

20. zordoloom

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

21. zordoloom

22. nbedwards934

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

Yep