A Math Problem for NS

JustDidIt

JustDidIt

Active member
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
 
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.
 
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)
 
=> 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
 
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