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A Math Problem for NS

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JustDidIt
Posts: 2194 - Karma: 160
Yep, it's a homework question, ignore if you don't want to solve. I've tried it a few different ways with no luck. It's for extra credit which I dearly need.
+k for help


Prove that:

(cosx / (1 - tanx))+(sinx /( 1 - cotx)) = sinx + cosx
Jan 24 2011 7:11PM
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WVskiing
Posts: 3007 - Karma: 1,949
http://lmgtfy.com/?q=(cosx+%2F+(1+-+tanx))%2B(sinx+%2F(+1+-+cotx))+%3D+sinx+%2B+cosx+
Jan 24 2011 7:14PM
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PJS.
Posts: 11943 - Karma: 895
i just pooped a lil
Jan 24 2011 7:14PM
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mommy
Posts: 12072 - Karma: 59
its 10
Jan 24 2011 7:15PM
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gabe.
Posts: 1891 - Karma: 47
google it. theres a bunch of sites that answer it. i personally have no idea what this shit is

http://lmgtfy.com/?q=Prove+that%3A+%20+(cosx+%2F+(1+-+tanx))%2B(sinx+%2F(+1+-+cotx))+%3D+sinx+%2B+cosx
Jan 24 2011 7:15PM
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SkiThe603
Posts: 5376 - Karma: 498
i could actually figure it out if i wanted to but i have to write a paper so it ain't happening
Jan 24 2011 7:15PM
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Talacious[Tally]
Posts: 1686 - Karma: 189
substitution bro
Jan 24 2011 7:16PM
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95959
Posts: 2765 - Karma: 377
well, tanx=sinx/cosx so you have (cosx)/1-(sinx/cosx). multiply by the reciprocal and the cosx goes away leaving you with sinx/1. same method on the other one should give you cos x.

Disclaimer: i could be wrong I only glanced at the problem.
Jan 24 2011 7:17PM
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95959
Posts: 2765 - Karma: 377
im wrong
Jan 24 2011 7:22PM
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Heart
Posts: 6156 - Karma: 3,462
here you go, straight from yahoo answers:

I think you mean (cos x/1-tan x)+(sin x/1-cot x)=sin x + cos x

Replace tan x with sin x/cos x , and replace cot x with cos x/sin x :

(cos x)/(1 - tan x) + (sin x)/(1- cot x) = (cos x)/(1 - ((sin x)/cos x)) + (sin x)(1 - ((cos x)/sin x))

= (cos² x)/(cos x - sin x) + (sin² x)/(sin x - cos x)
= cos² x - sin² x)/(cos x - sin x)
= (cos x + sin x)(cos x - sin x)/(cos x - sin x)
= cos x + sin x provided cos x - sin x ≠ 0 (in which case the denominators of the original expression aren't defined anyway)
Jan 24 2011 7:29PM
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JustDidIt
Posts: 2194 - Karma: 160
I'm not sure why, the yahoo answers answer isn't doing it for me, unless i'm just missing it. i hate math
Jan 24 2011 7:36PM
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SketchySwiss
Posts: 393 - Karma: 43
=> cosx+ (sinx/(1-cotx)*(1-tanx)= sinx + cosx - sinxtanx - cosxtanx

=> cosx(1-cotx) + sinx(1-tanx) = (sinx + cosx - sinxtanx - cosxtanx)(1-cotx)

=> cosx + sinx -cosxcotx -sinxtanx = sinx + cosx -sinxtanx -cosxtanx -sinxcotx -cosxcotx +sinxtanxcotx + cosxtanxcotx

=> cosxtanx + sinxcotx = sinxtanxcotx + cosxtanxcotx

=> sinx + cosx = sinx + cosx

there you are. I m pretty fucked (2am) but I m pretty sure this is it. I just used the trigonometrical identities
Jan 24 2011 7:40PM
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