#Inverse trigo
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What's in the input for arctan?
its from -3 to 3
...no.
Because "from -3 to 3" isn't the expression inside of arctan.
Inverse tangent, also known as arctangent.
That is not the expression inside of arctan.
im talkkin about y
I know you are. I don't know why, because I'm not.
Of arctan.
reciprocal of that
Right. So what does that mean?
need to make cases...
They both belonga to different range so i guess we need to make cases
before converting cot to tan
Okay. So make the cases.
I did
And what are they? In terms of 6y/(9 - y^2)?
like
when
y is +ve
we can reci[rpcal that
but if -ve
oh wait im confused over -ve
its gonna pie + ?
Look. Let x = 6y/(9 - y^2). Then what we have is arctan(x) + arccot(1/x).
right
So you're correct that we split it into cases x > 0 and x < 0.
yeah
Let's consider - wait, have you studied calculus at all?
yes
Okay, so I can talk about limits.
yeah
What's the limit as x approaches 0 from below of arccot(1/x)?
No.
okay
The limit as x approaches 0 from below of 1/x is negative infinity, right?
wait
you sure
wont it depend
ig thats infinity
like shouldnt that be 0-
I have to go rn gonna come tomorrow again thanks for the help tho
...I said that. 0 from below.
yea that should be -ve infinity
Right. So it's the limit as u goes to minus infinity of arccot(u).