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.

juriemagic May 6, 2021

#1**+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.

=^._.^=

catmg May 6, 2021

#1**+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

#3**+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**+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**+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