Here's the question you clicked on:

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

• This Question is Closed
1. Atkinsoha Group Title

2. getusel Group Title

$\frac{ cosx }{ sinx }$

3. Atkinsoha Group Title

Can you show me how to figure that out?

4. getusel Group Title

sure

5. getusel Group Title

sorry

6. getusel Group Title

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

7. getusel Group Title

is this the question

8. getusel Group Title

because in the picture the question does not seem right

9. getusel Group Title

if this is the question

10. getusel Group Title

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

11. getusel Group Title

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

12. getusel Group Title

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

13. getusel Group Title

cosx/sinx=cotx=1/tanx

14. getusel Group Title

done

15. Atkinsoha Group Title

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

16. getusel Group Title

for sure

17. Atkinsoha Group Title

okay, hold on let me get it..

18. Atkinsoha Group Title

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

19. getusel Group Title

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

20. getusel Group Title

Here is the final answer

21. Atkinsoha Group Title

So what's the final answer?

22. getusel Group Title

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

23. getusel Group Title

does it make sense

24. Atkinsoha Group Title

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

25. getusel Group Title

If you practice, it is not that hard my friend

26. Atkinsoha Group Title

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

27. getusel Group Title

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

28. Atkinsoha Group Title

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

29. getusel Group Title

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

30. getusel Group Title

it is verified

31. Atkinsoha Group Title

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

32. getusel Group Title

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

33. Atkinsoha Group Title

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

34. getusel Group Title

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 Group Title

ohh okay. thanks.

36. getusel Group Title

it is my pleasure