+1124 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.
+2401 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.
=^._.^=
+2401 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.
=^._.^=
+1124 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
+1124 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...
+2401 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.
=^._.^=
