Find the exact value of tan inverse

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Find the exact value of tan inverse

Mathematics
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tan theta = opp / adj
(there is no inverse)

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this is a dumb question.. but what does the theta sign look like?
theta : \(\theta\)
that's the sign I used but I thought it meant inverse :0
\[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]\[\Downarrow\]\[\theta = \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)=\arctan\left(\frac{\text{opposite}}{\text{adjacent}}\right)\]
the question is only asking you to find \(\tan\theta\), you don't have to compute the value of theta,
It's saying the equation has too many equal signs
what is the length of the side opposite to theta in the diagram?
is 6 the opp and 2 and 5 the adj?
yes, opp = 6 and adj = 2√5
my calculator said it was 1/2 as the answer, but it said that 1/2 is not correct
\(\tan\theta = \dfrac{6}{2\sqrt 5}\)
now cancel the common factor
what's the common factor?
\[\tan\theta = \dfrac{6}{2\sqrt 5}=\dfrac{2\times3}{2\times\sqrt 5}=\]
\[\tan \emptyset \frac{ 6}{ 5 }=\frac{ 3 }{ 5 }\]
now rationalize the denominator \[\tan\theta = \frac{6}{2\sqrt 5}=\frac{3}{\sqrt 5}=\frac{3}{\sqrt 5}\times\frac{\sqrt5}{\sqrt5}=\]
20..?
\[\frac{3}{\sqrt 5}\times\frac{\sqrt5}{\sqrt5}=\frac{3\times\sqrt5}{\sqrt5\times\sqrt5}\] now simplify the denominator
would it be 1 because 5 can go into 5 once?
\[\sqrt5\sqrt5=\sqrt5^2= .....\]
i'm not even sure how to go about that :o
\[√x^2 = x\]
oh gosh
ikr
my options don't even make sense then
we are very close to one of those
what is the square root of five squared?
5? I think
oh wait i did that wrong
yes, so what is our final answer?
5 squared over
3 I wanna say

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