## anonymous 4 years ago combine the fractions and simplify to a multiple of a power of a basic trigonometric function sinx/cot^2x-sinx/cos^2x?

1. anonymous

can i change the sinx/cos^2x into tan^2x?

2. zepdrix

Hmm no. You can take ONE of the cosines, and change it to tanx*(1/cosx) But I dont know if that's the right way to go c: hmm

3. zepdrix

$\large \frac{ \sin x }{ \cot^2 x }-\frac{ \sin x }{ \cos^2 x }=\frac{ \sin x }{ (\frac{ \cos^2 x }{ \sin^2 x }) }-\frac{ \sin x }{ \cos^2 x }$ $\large \frac{ \sin x }{ 1 }*\frac{ \sin^2 x }{ \cos^2 x }-\frac{ \sin x }{ \cos^2 x }$

4. zepdrix

There are multiple ways to simplify it, since there are so many trig identities, so you might find a different way... but this is what I was thinking :3

5. zepdrix

$\large \frac{ \sin^3 x - \sin x }{ \cos^2 x }$

6. zepdrix

Confused about any of that so far? :D

7. anonymous

how did you get sinx/1∗sin^2x/cos^2x

8. zepdrix

$\large \frac{ \left(\frac{ a }{ b }\right) }{ \left(\frac{ c }{ d }\right) }=\frac{ a }{ b }*\frac{ d }{ c }$

9. zepdrix

Remember how you can do that? when dividing fractions :D flip the bottom one and rewrite it as multiplication.

10. anonymous

but the top wasn't a fraction yet it was sinx/ (cos^2x/sin^2x)

11. anonymous

oh wait you put it over 1 to make it into a fraction huh?

12. zepdrix

$\huge \frac{ \left(\sin x \right) }{ \left(\frac{ \cos^2 x }{ \sin^2 x } \right) }=\frac{ \left(\frac{ \sin x }{ 1 } \right) }{ \left(\frac{ \cos^2 x }{ \sin^2 x } \right) }$

13. zepdrix

Yah hehe :3

14. zepdrix

I gotta go, so lemme post the rest of the steps, and you can at least look at them while I'm gone. I won't be here to answer questions though :c

15. anonymous

okay

16. zepdrix

$\large \frac{ \sin^3 x - \sin x }{ \cos^2 x }=\frac{ \sin x(\sin^2 x - 1) }{ 1-\sin^2 x }$

17. zepdrix

Factored a sinx from each term in the top. Rewrote cos^2 as 1-sin^2 using identity :D

18. zepdrix

Now if we factor a -1 ouch of each term in the top brackets, we'll have a nice easy cancellation.

19. zepdrix

$\large \frac{ -\sin x(1-\sin^2 x) }{ 1-\sin^2 x }=-\sin x$

20. zepdrix

Hopefully I didn't make a mistake in there :D heh

21. anonymous

alright well thanks