## anonymous 4 years ago help with secx/sinx-sinx/cosx!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1. anonymous

you wanna simplify it? note that sec x = 1/cos x so it becomes $$\frac{1}{sin x cos x} - \frac{sin x}{cos x}$$ if you continue to put both fractions into one , you'll notice something beautiful.

2. anonymous

well i got ((1/cosx)/sinx)-sinx/cos x and IDK what what to do after that unless i change cosx to 1/secx then can i cancel out the two secx?

3. anonymous

The sec x cannot be "cancelled". from above, $\frac{ 1 }{ \sin x \cos x }-\frac{ \sin x }{ \cos x }=\frac{ 1-\sin x(\sin x) }{ \sin x \cos x}$ and $$cos^2 x+sin^2 x =1$$ :) I think this will help.

4. anonymous

but if you subtract sin from 1 how did you get 1-sinx(sinx)? wouldn't it just be 1-sinx?

5. anonymous

hmm...I didn't subtract sin x.. $\frac{ 1 }{ a }-\frac{ 1 }{ b }=\frac{ b-a }{ ab }$ I think this should clear things up.

6. anonymous

but you have two things in the first one cuz it's1/sinxcosx not just 1/sinx

7. anonymous

yup! :) so I multiplied sin x to $$\frac {sin x}{cos x}$$ to have the same denominator.

8. anonymous

oh okay then you seperated it into 1-sinx(sinx) because when you multiply it you get 1-sin^2x

9. anonymous

but now what? cuz i know it has to equal cotx but i don't see how

10. anonymous

well, since from $$cos^2x+sin^2x=1$$ we know that $$cos^2 x=1-sin^2 x$$ so it simplifies to $$\frac{cos^2}{sin x cos x}$$=cot x

11. anonymous

but there's a cosx in the denominator so how does that work if cot x =cosx/sinx?

12. anonymous

well, $$\frac{cos^2 x}{ sinxcosx}$$= $$\frac{cos x cosx}{sin x cos x}$$ and we can cancel the cos x from the numerator and denominator.

13. anonymous

oh okay!!!! weLl thanx

14. anonymous

You're welcome :)