## kaylala one year ago topic: PROVING IDENTITIES - help me please (SEE COMMENTS) (trigonometry)

1. kaylala

@Loser66 help

2. SolomonZelman

3. SolomonZelman

( Loser66 is offline )

4. kaylala

(1)/(sec x - tan x) = sec x + tan x

5. kaylala

i noticed

6. SolomonZelman

$\huge\color{blue}{ \frac{1}{\sec(x)-\tan(x)} =\sec(x)+\tan(x) }$

7. SolomonZelman

$\huge\color{blue}{ 1 =(\sec(x)+\tan(x) )(\sec(x)-\tan(x))}$

8. SolomonZelman

I multiplied both sides times sec(40)-tan(40)

9. kaylala

why? is that possible? i thought we can only manipulate 1 side?

10. kaylala

shouldn't the other side be stable?

11. SolomonZelman

Oh, then I take what I wrote back.

12. kaylala

okay

13. SolomonZelman

$\huge\color{blue}{ \frac{1}{\sec(x)-\tan(x)}=\sec(x)+\tan(x) }$ LEFT SIDE:\ top and bottom times sec(x)+tan(x) $\huge\color{blue}{ \frac{\sec(x)-\tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) }$ Now, use the identity for the bottom, $\sec^2x=\tan^2x+1$

14. SolomonZelman

Oh, the top should say + not - sorry.

15. SolomonZelman

$\huge\color{blue}{ \frac{\sec(x) \huge\color{red}{ + } \tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) }$

16. SolomonZelman

HUGE HINT: $\sec^2x=\tan^2x+1~~~~~~~ther efore~~~~~\sec^2x-\tan^2x=1$

17. kaylala

i do not get it. sorry

18. SolomonZelman

Do you get what I did before. the last huge post in blue with a red plus ?

19. kaylala

nope but hey i found this: http://symbolab.com/solver/trigonometric-identity-calculator/prove%20%5Cfrac%7B1%7D%7B%5Csec(x)-%5Ctan(x)%7D%3D%5Csec(x)%2B%5Ctan(x) but the answer doesnt seem to be equal is there a way you could make it equal???

20. SolomonZelman

Don't try to look it up else where, follow me. lets start over, ok? $\huge\color{green} { \frac{1}{\sec(x)-\tan(x)}=\sec(x)+\tan(x) }$ I am working the left side: 1) Multiply top and bottom times sec(x)+tan(x)

21. kaylala

ok

22. SolomonZelman

$\color{green} { \frac{\sec(x)+\tan(x)}{(~~\sec(x)-\tan(x)~~)~(~~\sec(x)+\tan(x)~~)}=\sec(x)+\tan(x) }$ $\color{green} { \frac{\sec(x)+\tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) }$ tell me if you don't get it.

23. kaylala

okay

24. SolomonZelman

Now lets use an identity. $\tan^2(x)+1=\sec^2(x)$ let's draw some conclusions. $\tan^2(x)+1\color{blue} { -\tan^2(x) } =\sec^2x\color{blue} { -\tan^2(x) }$$1=\sec^2x-\tan^2x$ look at the denominator, what is it now equal to? has your identity been verified?

25. kaylala

yes. got it thanks @SolomonZelman

26. SolomonZelman

You welcome, no problem, just knowing identities that's all the skill involved.

27. kaylala

and wow. that was shorter

28. SolomonZelman

Yeah, I explained the same thing, but in a shorter way.

29. kaylala

what's the technique?

30. kaylala

how come you're so good at this?

31. SolomonZelman

what do you mean, the technique for what? Ohhh, I took trig last year, and got 97. I am just good at math and bad at other staff.

32. kaylala

technique in answering / proving these trigonometric identities? you did it so fast and in a shorter and much efficient way that's really cool WOW! Congratz! good for you

33. SolomonZelman

Thank you! I am sure that if you try to do a lot of practice problems you will master them.

34. kaylala

i really hope so. failing trigonometry is the last thing that i'd want

35. SolomonZelman

36. kaylala

last semester i got 2.0 in algebra now i'm taking trigonometry and it do is very hard. i dont have my grade yet since the 2nd semester just started.

37. kaylala

1 being the highest 3 - pass 5 - fail

38. kaylala

@SolomonZelman ^