Question

Evaluate arccot(10/23) in radians and round to 2 decimals.
I got this wrong when I entered 1/tan-1(10/23)=2.44 what did i do wrong?

Answer 1

\[\cot^{-1}(x) \neq \frac{1}{\tan^{-1}(x)} \]

Answer 2

that's what I used i think but the professor said my answer was wrong?

Answer 3

To see this, let
\[y = \cot^{-1}(x) \rightarrow \cot(y) = x\]
Then
\[\tan(y) = \frac{1}{x} \]
and so
\[ y = \tan^{-1}\left(\frac{1}{x}\right) \]
Which means that
\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]

Answer 4

No, it's NOT equal. The slash through the equals sign means not equal.

Answer 5

I'm sorry, I'm tired. so my question would be tan^-1(1/(10/23)) ???

Answer 6

Yes. While it's true that
\[\cot(x) = \frac{1}{\tan(x)} \]
the same is NOT true for the inverse functions. Instead, as shown above,
\[\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]
so
\[\cot^{-1}\left(\frac{10}{23}\right) = \tan^{-1}\left(\frac{23}{10}\right) \approx 1.16 \]

Answer 7

oh, thank you so much! I kept getting these problems wrong as one person had me working them the other way.

Answer 8

I am sorry to hear that. But, I suppose everybody makes mistakes, even people who are trying to help :)

Answer 9

it makes sense now. thank you!