## anonymous 3 years ago Verify, then find which is NOT equivalent : (Picture below)

1. anonymous

2. anonymous

$\frac{ cosx }{ sinx }$

3. anonymous

Can you show me how to figure that out?

4. anonymous

sure

5. anonymous

sorry

6. anonymous

$\sin(\pi/2-x)\csc$

7. anonymous

is this the question

8. anonymous

because in the picture the question does not seem right

9. anonymous

if this is the question

10. anonymous

$\sin(\frac{ \pi }{2 }-x)=\sin(\frac{ \pi }{ 2 })cosx-\cos(\frac{ \pi }{ 2 })sinx=1.cosx-0.sinx=cosx$

11. anonymous

having this in mind, let us insert it in the original expression

12. anonymous

$\sin(\frac{ \pi }{2 }-x)cscx=cosxcscx=cosx.\frac{ 1 }{ sinx }$

13. anonymous

cosx/sinx=cotx=1/tanx

14. anonymous

done

15. anonymous

Thank you! I have another one that maybe you could help me with?

16. anonymous

for sure

17. anonymous

okay, hold on let me get it..

18. anonymous

If you could do it all in one message that would be great so I can follow along easier.

19. anonymous

$\frac{ cscA-sinA }{ \cot ^{2}A }=\frac{\frac{ 1 }{sinA }-sinA }{ \frac{ \cos ^{2}A }{ \sin ^{2}A } }=\frac{ \frac{ 1-\sin ^{2}A }{sinA } }{ \frac{ \cos ^{2}A }{\sin ^{2} A} }$ $\frac{ 1-\sin ^{2}A }{sinA }.\frac{ \sin ^{2} A}{ \cos ^{2}A }=\frac{ \cos ^{2}A sinA}{ cosA }=sinAcosA=\frac{ 1 }{2 }\sin(2A)$

20. anonymous

21. anonymous

22. anonymous

$\frac{ 1 }{ 2 }\sin(2A)$

23. anonymous

does it make sense

24. anonymous

uhmm.. yeah sort of. im terrible at this whole simplifying thing. up for a few more?

25. anonymous

If you practice, it is not that hard my friend

26. anonymous

i got all of the verifying ones but this one..

27. anonymous

I am going for launch. If you stay for 30 minutes i will get back to you.

28. anonymous

I'll be here all evening. Thank you! Have a good time :)

29. anonymous

$\frac{ \cos \alpha-\cos \beta }{ \sin \alpha+\sin \beta }+\frac{ \sin \alpha-\sin \beta }{ \cos \alpha+\cos \beta}=$ $\frac{ \cos ^{2}\alpha-\cos ^{2}\beta+\sin ^{2}\alpha-\sin^2\beta }{(\sin \alpha+\sin \beta)(\cos \alpha +\cos \beta)}$ =$\frac{ \cos^2 \alpha+\sin^2\alpha-(\sin^2\beta +\cos^2\beta) }{(\sin \alpha+\sin \beta)(\cos \alpha+\cos \beta) }=\frac{ 1-1 }{(\sin \alpha+\sin \beta) (\cos \alpha+\cos \beta) }=0$

30. anonymous

it is verified

31. anonymous

how is the denominator verified? nothing was done to it?

32. anonymous

since it does not have any significance to simplify the denominator, we can leave it as it is.

33. anonymous

How do you know if it has significance to simplify it or not?

34. anonymous

In this particular problem, the purpose is to simplify the right hand side to zero. That tells us the numerator of the RHS has to be siplified to zero. our focus should be mostly on working out the numerator.

35. anonymous

ohh okay. thanks.

36. anonymous

it is my pleasure