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## kaylala Group Title topic: PROVING IDENTITIES - help me please (SEE COMMENTS) (trigonometry) 6 months ago 6 months ago

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1. kaylala Group Title

@Loser66 help

2. SolomonZelman Group Title

post your question please.

3. SolomonZelman Group Title

( Loser66 is offline )

4. kaylala Group Title

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

5. kaylala Group Title

i noticed

6. SolomonZelman Group Title

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

7. SolomonZelman Group Title

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

8. SolomonZelman Group Title

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

9. kaylala Group Title

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

10. kaylala Group Title

shouldn't the other side be stable?

11. SolomonZelman Group Title

Oh, then I take what I wrote back.

12. kaylala Group Title

okay

13. SolomonZelman Group Title

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

Oh, the top should say + not - sorry.

15. SolomonZelman Group Title

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

16. SolomonZelman Group Title

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

17. kaylala Group Title

i do not get it. sorry

18. SolomonZelman Group Title

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

19. kaylala Group Title

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 Group Title

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 Group Title

ok

22. SolomonZelman Group Title

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

23. kaylala Group Title

okay

24. SolomonZelman Group Title

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 Group Title

yes. got it thanks @SolomonZelman

26. SolomonZelman Group Title

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

27. kaylala Group Title

and wow. that was shorter

28. SolomonZelman Group Title

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

29. kaylala Group Title

what's the technique?

30. kaylala Group Title

how come you're so good at this?

31. SolomonZelman Group Title

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 Group Title

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 Group Title

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

34. kaylala Group Title

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

35. SolomonZelman Group Title

WHAT?! FAILING?! what is your current grade may I ask ?

36. kaylala Group Title

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 Group Title

1 being the highest 3 - pass 5 - fail

38. kaylala Group Title

@SolomonZelman ^