How do you calculate $ {\cot ^{ - 1}}0.684? $
Answer
630.6k+ views
Hint: Make sure what $ {\cot ^{ - 1}}0.684 $ actually means, if $ - 1 $ is indicating the power of the given function or the inverse function.
If it is in the power of the function then just change $ \cot $ to $ \tan $ and find the required value.
But if, $ - 1 $ is showing the inverse function then use inverse trigonometry in order to solve this problem.
Complete step-by-step answer:
The solution for given question have two possibilities-
If $ {\cot ^{ - 1}}0.684 $ means the power of $ \csc 0.5 $ is $ - 1 $
And if $ {\csc ^{ - 1}}0.5 $ it is an inverse trigonometric function.
So let us go with the first possibility, that is $ - 1 $ in $ {\csc ^{ - 1}}0.5 $ is indicating the power of $ \csc 0.684 $
Power of a function $ - 1 $ means the following:
$ {f^{ - 1}}(x) = \dfrac{1}{{f(x)}} $
So doing this with given function we will get
$ {\cot ^{ - 1}}0.684 = \dfrac{1}{{\cot 0.684}} $
Now we all know that $ \tan x = \dfrac{1}{{\cot x}} $ , so substituting it above, we will get
$ \dfrac{1}{{\cot 0.684}} = \tan 0.684 $
Here find the value of $ \tan 0.684 $ in your calculator you will get a value close to \[0.8153..\]
Now solving the second possibility, that is $ {\cot ^{ - 1}}0.684 $ is an inverse trigonometric function
Let us assume, $ {\cot ^{ - 1}}0.684 = x $ then
We can write $ 0.684 = \cot x $
Now we know that $ \tan x = \dfrac{1}{{\cot x}} $ therefore we can write it as
$
0.684 = \dfrac{1}{{\tan x}} \\
\Rightarrow \tan x = 1.462 \\
\Rightarrow x = {\tan ^{ - 1}}(1.462) \;
$
We have to calculate this value using a calculator, to calculate the inverse function press shift key on your calculator and then press $ \tan $ and put the value, the answer will come close to $ 0.970893251 $ in radians and $ {55.6^ \circ } $ in degrees.
So, the correct answer is “ $ {55.6^ \circ } $ ”.
Note: Generally inverse trigonometric functions have prefix arc written before them to indicate that it is an inverse trigonometric function, e.g. $ \arcsin ,\;\arccos $
In this question the argument of $ \cot $ is written in radians, so keep this in mind when doing the calculations.
If it is in the power of the function then just change $ \cot $ to $ \tan $ and find the required value.
But if, $ - 1 $ is showing the inverse function then use inverse trigonometry in order to solve this problem.
Complete step-by-step answer:
The solution for given question have two possibilities-
If $ {\cot ^{ - 1}}0.684 $ means the power of $ \csc 0.5 $ is $ - 1 $
And if $ {\csc ^{ - 1}}0.5 $ it is an inverse trigonometric function.
So let us go with the first possibility, that is $ - 1 $ in $ {\csc ^{ - 1}}0.5 $ is indicating the power of $ \csc 0.684 $
Power of a function $ - 1 $ means the following:
$ {f^{ - 1}}(x) = \dfrac{1}{{f(x)}} $
So doing this with given function we will get
$ {\cot ^{ - 1}}0.684 = \dfrac{1}{{\cot 0.684}} $
Now we all know that $ \tan x = \dfrac{1}{{\cot x}} $ , so substituting it above, we will get
$ \dfrac{1}{{\cot 0.684}} = \tan 0.684 $
Here find the value of $ \tan 0.684 $ in your calculator you will get a value close to \[0.8153..\]
Now solving the second possibility, that is $ {\cot ^{ - 1}}0.684 $ is an inverse trigonometric function
Let us assume, $ {\cot ^{ - 1}}0.684 = x $ then
We can write $ 0.684 = \cot x $
Now we know that $ \tan x = \dfrac{1}{{\cot x}} $ therefore we can write it as
$
0.684 = \dfrac{1}{{\tan x}} \\
\Rightarrow \tan x = 1.462 \\
\Rightarrow x = {\tan ^{ - 1}}(1.462) \;
$
We have to calculate this value using a calculator, to calculate the inverse function press shift key on your calculator and then press $ \tan $ and put the value, the answer will come close to $ 0.970893251 $ in radians and $ {55.6^ \circ } $ in degrees.
So, the correct answer is “ $ {55.6^ \circ } $ ”.
Note: Generally inverse trigonometric functions have prefix arc written before them to indicate that it is an inverse trigonometric function, e.g. $ \arcsin ,\;\arccos $
In this question the argument of $ \cot $ is written in radians, so keep this in mind when doing the calculations.
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