## anonymous 3 years ago cot(theta)-tan(theta) over sin(theta) + cos(theta) equals csc(theta) - sec(theta) how do I proof this?

• This Question is Open
1. Mertsj

Let's use x instead of theta

2. anonymous

okay

3. Mertsj

$\frac{\cot x-\tan x}{\sin x+\cos x}=\csc x-\sec x$

4. Mertsj

Do I have the problem stated correctly?

5. anonymous

yes :)

6. anonymous

would I have to replace cot(x) with cos(x)/sin(x)

7. anonymous

and tan(x) with sin(x)/cos(x)

8. Mertsj

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

9. Mertsj

Can you get it from there?

10. anonymous

sadly no? can you explain how I can simplify it more please?

11. Mertsj

Factor the numerator and then the cosx + sinx will cancel.

12. anonymous

I'm still lost :(

13. Mertsj

@alejandrop95 This is not your question

14. anonymous

how would I factor it though? I'm sorry I am new to trig.

15. Mertsj

Can you factor this: $a^2-b^2$

16. anonymous

I don't think you can because it doesn't have the same coefficient

17. Mertsj

$a^2−b^2=(a-b)(a+b)$

18. Mertsj

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

19. anonymous

oh okay that makes sense. so would it be cos-sin(cos+sin)

20. Mertsj

yes

21. anonymous

then that would cancel with the denominator correct?

22. Mertsj

Yes. sinx + cosx will cancel

23. anonymous

which will leave 1/sin(x)+cos(x)

24. Mertsj

no

25. anonymous

using the reciprocal identities this will equal csc(x)-sec(x) ?

26. Mertsj

|dw:1382579245755:dw|

27. anonymous

cos(x)-sin(x)/sin(x)+cos(x0

28. anonymous

this is where I would use the reciprocal identities?

29. Mertsj

|dw:1382579370642:dw|

30. anonymous

the cos(x) cancel and the sin(x) cancels?

31. Mertsj

|dw:1382579509266:dw|

32. anonymous

yay!!!! the cancel and its one,which now you are left with the csc(x) and sec(x) reciprocal identities!! Thank you!! the key idea here that I need to work on was the factoring

33. anonymous

thank you !:)

34. anonymous

thanks ! :)

35. Mertsj

yw