## Atkinsoha Group Title Verify, then find which is NOT equivalent : (Picture below) one year ago one year ago

1. Atkinsoha

2. getusel

$\frac{ cosx }{ sinx }$

3. Atkinsoha

Can you show me how to figure that out?

4. getusel

sure

5. getusel

sorry

6. getusel

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

7. getusel

is this the question

8. getusel

because in the picture the question does not seem right

9. getusel

if this is the question

10. getusel

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

11. getusel

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

12. getusel

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

13. getusel

cosx/sinx=cotx=1/tanx

14. getusel

done

15. Atkinsoha

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

16. getusel

for sure

17. Atkinsoha

okay, hold on let me get it..

18. Atkinsoha

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

19. getusel

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

20. getusel

21. Atkinsoha

22. getusel

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

23. getusel

does it make sense

24. Atkinsoha

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

25. getusel

If you practice, it is not that hard my friend

26. Atkinsoha

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

27. getusel

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

28. Atkinsoha

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

29. getusel

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

30. getusel

it is verified

31. Atkinsoha

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

32. getusel

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

33. Atkinsoha

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

34. getusel

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. Atkinsoha

ohh okay. thanks.

36. getusel

it is my pleasure