Thread: trig identity

Hey
Prove: cos(x)/ 1-tan(x) + sin(x)/ 1-cot(x) = sin(x) + cos(x)

i subbed 1-tan(x)= sin(x)/cos(x)
and cot(x) = cos(x)/sin(x)

so i divided it and times it all out

and i get to this stage

1/ 2cos(x)sin(x)-cos^2(x)-sin^2(x) and didnt knwo what to do

Thanks

Hi smmmc

Originally Posted by smmmc

i subbed 1-tan(x)= sin(x)/cos(x)

You forgot the term "1"

$\displaystyle 1-\tan(x) = 1-\frac{\sin(x)}{\cos(x)}$

yerh i did that.
Let me just show you my whole working out. sorry its not in latex, dont know how to use it

So,

cos(x)/(1- sin(x)/cos(x)) + sin(x)/(1-cos(x)/sin(x))

= cos(x)/(cos(x)-sin(x)/cos(x)) + sin(x)/(sin(x)-cos(x)/sin(x))

= cos(x) * cos(x)/(cos(x)-sin(x)) + sin(x) * (sin(x)/(sin(x)-cos(x))

= cos^2(x) + sin^2(x)/ (cos(x)-sin(x)*(sin(x)/(sin(x)-cos(x))

which equals

1/ (2cos(x)sin(x)-cos^2(x)-sin^2(x))

help please

Hi smmmc

Originally Posted by smmmc

cos^2(x) + sin^2(x)/ (cos(x)-sin(x)*(sin(x)/(sin(x)-cos(x))

What's this ??

cos(x) * cos(x)/(cos(x)-sin(x)) + sin(x) * (sin(x)/(sin(x)-cos(x))

It's good till this part. Then, you can manipulate it like this :

$\displaystyle \frac{\cos^{2}(x)}{\cos(x)-sin(x)}+\frac{sin^{2}(x)}{\sin(x) - cos(x)}$

$\displaystyle = \frac{\cos^{2}(x)}{\cos(x)-sin(x)} - \frac{sin^{2}(x)}{\cos(x) - sin(x)}$

And a little suggestion from me : How about learning latex ? http://www.mathhelpforum.com/math-he...-tutorial.html

so it then becomes

cos^2(x)-sin^2(x)/ cox(x)-sin(x)

can you cross it out? simplify?

but that leads to cos(x)-sin(x).. and i dint know you could simplify like that?

help

Hi smmmc

$\displaystyle \frac{\cos^{2}(x)-sin^{2}(x)}{\cos(x)-sin(x)}$ doesn't simplify to $\displaystyle \cos(x) - \sin(x)$

$\displaystyle \cos^{2}(x)-sin^{2}(x) = (\cos x - \sin x) (\cos x+\sin x)$