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How do you evaluate \[\arctan \left( {\dfrac{1}{2}} \right)\]?

Answer
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Hint: Here the arctan function of any variable is defined as the inverse tangent function of that variable and it’s the other name of the inverse tangent function. We will substitute the value of the required inverse tangent function using the trigonometric table to get the required value.

Complete step by step solution:
 Here we need to find the value of the given trigonometric function i.e. \[\arctan \left( {\dfrac{1}{2}} \right)\].
We know that the arctan function is the other name of the inverse tangent function.
We can write \[\arctan \left( {\dfrac{1}{2}} \right)\] as \[{\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\].
Let the value of \[{\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\] be \[x\] i.e.
\[{\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) = x\]
Now, we will take a tangent function on both sides of the equation. Therefore, we get
 \[ \Rightarrow \tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right) = \tan x\]
We know from the properties of the inverse trigonometric function that \[\tan \left( {{{\tan }^{ - 1}}x} \right) = x\].
We will use the same property of the inverse trigonometric function. Therefore, we get
\[ \Rightarrow \dfrac{1}{2} = \tan x\]
Now, we will use the table to the tangent function to find the value of \[x\].
Using the data from the table, we get
  \[ \Rightarrow \tan \left( {26^\circ 33'} \right) = \tan x\]
Therefore, the value of \[x\] is equal to \[26^\circ 33'\].
As we have assumed that \[{\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) = x\] and we also know that \[\arctan \left( {\dfrac{1}{2}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\].

Hence, the value of \[\arctan \left( {\dfrac{1}{2}} \right)\] is equal to \[26^\circ 33'\].

Note: Trigonometric identities are defined as the equalities that include the basic trigonometric functions. They are always true for every value of the occurring variables where both sides of the equality are well defined. Inverse trigonometric functions are also known as the “Arc Functions” since, for a given value of trigonometric functions, they always produce the length of the arc which is needed to obtain that particular value. The inverse trigonometric functions perform the inverse operation of the trigonometric functions such as sine, cosine, tangent, secant, cosecant, and cotangent.