juriemagic

Hi mworkhard222,

I've got it...thank you very much..

Hi mworkhard222,

thank you for the quick response...but please just tell me if you don't mind...

I thought the factors were already out there...meaning the (x+2) and the (2x-3)??

Hi heureka,

Thank you for the response, however, could you kindly explain how you got to the rule in the first place...how did you know it had to be \({(x+1)\over2}*5\)

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...

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

Thank you catmg,

very helpfull, I do appreciate..

Hello dear Melody.....

Thank you for confirming....yes I know its the triangle, just do not know where to get the symbol...I searched but could not find it...

But then I did this: leave it as 

\(kx^2+x(4k-1)+4k-2\)

\(d=b^2-4ac\)

\(=(4k-1)^2-4(k)(4k-2)\)

\(=16k^2-8k+1-16k^2+8k\)

\(=1\)

So the roots are Real, un-equal and rational....is this right perhaps?

hi Alan,

I understand that thank you very much, my problem is how to deal with the 6pq?

Ok Alan, I understand...the 6pq is derived at when the \((p+3q)^2\) is calculated. so you convert the problem to a trinomial basically, and leave the 4th term on its own as a square, so when the trinomial portion is factored, you just add the 4th term and then end up with two terms again which are complete squares as well...then carry on from there to get the final answer...

Thank you very much Alan, I do appreciate..

hi dani0099,

Thank you, will check those out..