+0  
 
0
77
5
avatar+846 

Hi friends...I want to take this oportunity to apologise for not coming back the last few times I had been helped. Especially when I ask for help late at night, hoping to see a response the next morning, only then to forget about it because my mornings are sometimes just hectic getting the day started. Please except my apology. I need help with a sum that I have really tried and tried but off course just cant seem to solve. I need to give an answer to my pupil tomorrow morning, so I WILL come back to you. 

 

It is to prove \({cosx \over{1-tanx}}+{SinxTanx \over{Tanx-1}}=Sinx+Cosx\)

 

Please friends help me, I promise to come back to you in the morning....Thank you all so very much. It is always appreciated.

 May 6, 2021

Best Answer 

 #1
avatar+1973 
+1

sine(x)*tan(x)-cos(x)

_______________  = sine(x) + cos (x)      # I'm using that line as division cause it's easier to visualize I think 

tan(x) - 1

 

sine(x)*sine(x)/cos(x)-cos(x)

_______________                     

sine(x)/cos(x) - 1

 

sine^2(x)-cos^2(x)/cos(x)

_______________                     

(sine(x)-cos(x))/cos(x)

 

sine^2(x)-cos^2(x)

_______________

sine(x)-cos(x)

 

sine(x) = a

cos(x) = b

(a^2-b^2)/(a-b) = a + b

So our answer simplifies to sine(x) + cos(x)

 

Last time I tried your problem, Melody taught me this cool triangle trick that I would suggest you to go see, if you haven't yet. :))

Quite useful. 

=^._.^=

 May 6, 2021
 #1
avatar+1973 
+1
Best Answer

sine(x)*tan(x)-cos(x)

_______________  = sine(x) + cos (x)      # I'm using that line as division cause it's easier to visualize I think 

tan(x) - 1

 

sine(x)*sine(x)/cos(x)-cos(x)

_______________                     

sine(x)/cos(x) - 1

 

sine^2(x)-cos^2(x)/cos(x)

_______________                     

(sine(x)-cos(x))/cos(x)

 

sine^2(x)-cos^2(x)

_______________

sine(x)-cos(x)

 

sine(x) = a

cos(x) = b

(a^2-b^2)/(a-b) = a + b

So our answer simplifies to sine(x) + cos(x)

 

Last time I tried your problem, Melody taught me this cool triangle trick that I would suggest you to go see, if you haven't yet. :))

Quite useful. 

=^._.^=

catmg May 6, 2021
 #2
avatar+846 
+1

Thank you catmg,

 

very helpfull, I do appreciate..

juriemagic  May 7, 2021
 #3
avatar+846 
+1

Hi catmg

 

Actually, I'm a bit lost..I do not understand where you get the first line...

sine(x)*tan(x)-cos(x)

_______________  = sine(x) + cos (x)      

tan(x) - 1

juriemagic  May 7, 2021
 #4
avatar+846 
+1

this is how I do it..but get stuck..

First I swopped the two terms around

\({SinxTanx \over{Tanx-1}}+{Cosx \over{1-Tanx}}\)

 

\({SinxTanx \over{Tanx-1}}-{Cosx \over{Tanx-1}}\)

 

\({{(Sinx){Sinx \over{Cosx}}} \over{sinx \over{Cosx}}-1}-{Cosx \over{Sinx \over{Cosx}}}-1\)

 

\({{Sin^2x \over{cosx}} \over{Sinx \over{Cosx}}-1}-{Cosx \over{Sinx \over{Cosx}}}-1\)

 

\({{Sin^2x \over{Cosx}}*{Cosx \over{Sinx}}-1}-({{Cosx}*{Cosx \over{Sinx}}})-1\)

 

\((Sinx-1)-({Cos^2x \over{Sinx}})+1\)

 

\((Sinx)-({Cos^2x \over{Sinx}})\)

 

And I'm stuck...

juriemagic  May 7, 2021
 #5
avatar+1973 
+1

Sorry I'm a bit late, I was sleeping. :))

-cos(x)/(tan(x)-1) = cos(x)/(1-tan(x))  I simply multiplied both sides by -1, and since we now have the same denominator, as sine(x)*tan(x)/(tan(x)-1), we can just add the numerators together. 

As for your method, I think you messed up on the second to third line, you included the -1 outside of the denominator of the fraction. 

 

=^._.^=

catmg  May 7, 2021

22 Online Users

avatar
avatar