ITF simplification

Express $\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right),\ \ -\frac{3\pi}{2}<x<\frac{\pi}{2}$ in the simplest form.
58 Replies
TeXit
TeXit7mo ago
Nimboi
No description
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Nimboi
NimboiOP7mo ago
expressing sin(x) and cos(x) in their tan(x/2) forms, I got here: $\tan^{-1}\left(\frac{1}{1-\tan^{2}\left(\frac{x}{2}\right)}\right)$
TeXit
TeXit7mo ago
Nimboi
No description
Nimboi
NimboiOP7mo ago
what now?
BlindSniper
BlindSniper7mo ago
hmmm gimme a sec cos(θ)/1-sin(θ) can be expresssed as tan(π/4+θ/2)
Nimboi
NimboiOP7mo ago
you're correct
BlindSniper
BlindSniper7mo ago
bro what is that domain given
Nimboi
NimboiOP7mo ago
they used a different method to do that
BlindSniper
BlindSniper7mo ago
send
Nimboi
NimboiOP7mo ago
camera doesnt work ill just tell you
BlindSniper
BlindSniper7mo ago
I just have in my notes as standard result alr
Nimboi
NimboiOP7mo ago
cosx = sin(pi/2 - x) 1 - sinx = 1 - cos(pi/2 - x) sin(pi/2 - x) = 2sin(pi/4 - x /2)cos(pi/4 - x/2)
BlindSniper
BlindSniper7mo ago
oh so just derivation of it alr got it
Nimboi
NimboiOP7mo ago
1 - cos(pi/2 - x) = 2sin^2(pi/4 - x/2)
BlindSniper
BlindSniper7mo ago
yeah alr nice u got it right?
Nimboi
NimboiOP7mo ago
yeah but can i proceed from here cuz i thought of that first
BlindSniper
BlindSniper7mo ago
hmmm
Nimboi
NimboiOP7mo ago
didnt think to express cos(x) as sin(pi/2 - x) and whatnot
BlindSniper
BlindSniper7mo ago
yeah me neither keep in mind tho
Nimboi
NimboiOP7mo ago
bet anyways for boards cant use that standard result as a standard result
BlindSniper
BlindSniper7mo ago
true
Nimboi
NimboiOP7mo ago
@BlindSniper (BS) made a typo its 1 - tan(x/2) in the denom $\tan^{-1}\left(\frac{1}{1-\tan\left(\frac{x}{2}\right)}\right)$
TeXit
TeXit7mo ago
Nimboi
No description
BlindSniper
BlindSniper7mo ago
oh makes sense I wasn't getting what you got ok lemme see
Nimboi
NimboiOP7mo ago
@BlindSniper (BS) made a mistake along the way just start from the original problem lol main thoda stupid
BlindSniper
BlindSniper7mo ago
wow r u state board or cbse
Nimboi
NimboiOP7mo ago
cbse lmao havent done good math in a while
BlindSniper
BlindSniper7mo ago
then thoda problem to zaroor hai state board u can directly write hmmmmmm
Opt
Opt7mo ago
Yo what's going on?
Nimboi
NimboiOP7mo ago
ignore the silly in the middle look at the problem on top i wanna do it by expressing sin and cos as their tan(x/2) identities if that'll somehow work
BlindSniper
BlindSniper7mo ago
tried to reverse it but getting nowhere
No description
Opt
Opt7mo ago
Hmmm
Nimboi
NimboiOP7mo ago
i got it
BlindSniper
BlindSniper7mo ago
nice tan(x/2) in terms of sinx and cosx but you would need to know the answer for this
Nimboi
NimboiOP7mo ago
yeah you get here and then
BlindSniper
BlindSniper7mo ago
lite le baap
Nimboi
NimboiOP7mo ago
1 = tan(pi/4)
BlindSniper
BlindSniper7mo ago
yeah sure but .
Nimboi
NimboiOP7mo ago
yeah the answer is pi/4 + x/2 ah ok you mean
BlindSniper
BlindSniper7mo ago
I'm saying would you have known the answer if you didn't see the solution?
Nimboi
NimboiOP7mo ago
hard to think of that without knowing its of that form?
BlindSniper
BlindSniper7mo ago
yeah
Nimboi
NimboiOP7mo ago
if i knew i had to remove the tan then maybe yes this is a neat trick it seems more thinkable to me than expressing cosx as sin(pi/2 - x)
BlindSniper
BlindSniper7mo ago
yea true ig but it kind of reminds me of the complex number questions
Nimboi
NimboiOP7mo ago
which ones?
BlindSniper
BlindSniper7mo ago
whhere you have to convert in terms of cistheta lemme get one rq
BlindSniper
BlindSniper7mo ago
No description
BlindSniper
BlindSniper7mo ago
here you take it as sin(pi/2-x) pretty fire
Nimboi
NimboiOP7mo ago
man what the how does one have that level of foresight or is it more just screwing around
BlindSniper
BlindSniper7mo ago
nah but this is easy in this case but if you get directly in a trigonometry question I wouldn't think of it
Nimboi
NimboiOP7mo ago
💀
BlindSniper
BlindSniper7mo ago
if you see it that way
Nimboi
NimboiOP7mo ago
well, thanks dude ill mark as solved
BlindSniper
BlindSniper7mo ago
👍
Nimboi
NimboiOP7mo ago
+solved @BlindSniper (BS)
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