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nbedwards934
Group Title
tanθ+cotθ=cscθxsecθ
(I will draw what I have so far)
 one year ago
 one year ago
nbedwards934 Group Title
tanθ+cotθ=cscθxsecθ (I will draw what I have so far)
 one year ago
 one year ago

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nbedwards934 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354253316580:dw
 one year ago

jidery Group TitleBest ResponseYou've already chosen the best response.0
dw:1354253520763:dw
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
Are you trying to prove this identity??
 one year ago

nbedwards934 Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
Would you like further help?
 one year ago

nbedwards934 Group TitleBest ResponseYou've already chosen the best response.0
yes please
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
Very well. Let me start from the beginning them and go step by step.
 one year ago

nbedwards934 Group TitleBest ResponseYou've already chosen the best response.0
alright. Thank you
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.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)
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
Are you following so far?
 one year ago

nbedwards934 Group TitleBest ResponseYou've already chosen the best response.0
you're working with both sides? My professor told us to only work with one side and leave the other as is.
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
Alright, If you want, I can work only on one side.
 one year ago

nbedwards934 Group TitleBest ResponseYou've already chosen the best response.0
that would be better, thank you. I started work on the cscθxsecθ side
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
okay, would you mind if I work from the left side?
 one year ago

nbedwards934 Group TitleBest ResponseYou've already chosen the best response.0
I actually might have figured this one out. Give me just a minute to finish it
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
Does that help?
 one year ago

nbedwards934 Group TitleBest ResponseYou've already chosen the best response.0
I'll be honest, it's very difficult to interpret the equations you have typed up there, but I'm trying haha
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.1
Here. This might help a bit. It's solved by a computer.
 one year ago

nbedwards934 Group TitleBest ResponseYou've already chosen the best response.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!
 one year ago
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