Here's the question you clicked on:
kaylala
topic: PROVING IDENTITIES - help me please (SEE COMMENTS) (trigonometry)
post your question please.
( Loser66 is offline )
(1)/(sec x - tan x) = sec x + tan x
\[\huge\color{blue}{ \frac{1}{\sec(x)-\tan(x)} =\sec(x)+\tan(x) } \]
\[\huge\color{blue}{ 1 =(\sec(x)+\tan(x) )(\sec(x)-\tan(x))} \]
I multiplied both sides times sec(40)-tan(40)
why? is that possible? i thought we can only manipulate 1 side?
shouldn't the other side be stable?
Oh, then I take what I wrote back.
\[\huge\color{blue}{ \frac{1}{\sec(x)-\tan(x)}=\sec(x)+\tan(x) } \] 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, \[\sec^2x=\tan^2x+1\]
Oh, the top should say + not - sorry.
\[\huge\color{blue}{ \frac{\sec(x) \huge\color{red}{ + } \tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) } \]
HUGE HINT: \[\sec^2x=\tan^2x+1~~~~~~~ther efore~~~~~\sec^2x-\tan^2x=1\]
Do you get what I did before. the last huge post in blue with a red plus ?
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???
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)
\[\color{green} { \frac{\sec(x)+\tan(x)}{(~~\sec(x)-\tan(x)~~)~(~~\sec(x)+\tan(x)~~)}=\sec(x)+\tan(x) }\] \[\color{green} { \frac{\sec(x)+\tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) }\] tell me if you don't get it.
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?
yes. got it thanks @SolomonZelman
You welcome, no problem, just knowing identities that's all the skill involved.
and wow. that was shorter
Yeah, I explained the same thing, but in a shorter way.
how come you're so good at this?
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.
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
Thank you! I am sure that if you try to do a lot of practice problems you will master them.
i really hope so. failing trigonometry is the last thing that i'd want
WHAT?! FAILING?! what is your current grade may I ask ?
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.
1 being the highest 3 - pass 5 - fail