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**Hint:**Here in this question the given function is a function of inverse trigonometry function. To find the value of this function by using Maclaurin series or expansion, it formula can be defined as \[f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( {{x_0}} \right)}}{{n!}}} {\left( {x - {x_0}} \right)^n}\] further solve this formula for \[{\tan ^{ - 1}}\] on by simplification we get the required result.

**Complete step by step solution:**

The given function is an inverse trigonometry function. The basic inverse trigonometric functions are used to find the missing angles in right triangles.

To solve the given inverse trigonometry function by using a Maclaurin series or expansion.

Maclaurin series is a function that has an expansion series that gives the sum of derivatives of that function. The Maclaurin series of a function \[f\left( x \right)\] up to order n may be found using series \[\left[ {f,{\text{ }}x,{\text{ }}0,{\text{ }}n{\text{ }}} \right] \] .

The Maclaurin series is a special case of Taylor series when \[x = 0\] . The Maclaurin series is given by

\[f(x) = f({x_0}) + f'({x_0})\left( {x - {x_0}} \right) + \dfrac{{f''({x_0})}}{{2!}}{\left( {x - {x_0}} \right)^2} + \dfrac{{f'''({x_0})}}{{3!}}{\left( {x - {x_0}} \right)^3} + \,\, \cdot \cdot \cdot \]

In general formula of Maclaurin series is

\[\Rightarrow f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( {{x_0}} \right)}}{{n!}}} {\left( {x - {x_0}} \right)^n}\]

Where \[f'({x_0})\] , \[f''({x_0})\] , \[f'''({x_0})\] , … are the successive differentials when \[{x_0} = 0\] .

Consider the given inverse tan function \[{\tan ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\]

Before going to solve this, first find the Maclaurin series of \[{\tan ^{ - 1}}\left( x \right)\] when \[{x_0} = 0\] .

So let’s begin :

\[\Rightarrow f(x) = {\tan ^{ - 1}}\left( x \right)\]

\[\Rightarrow f\left( 0 \right) = 0\]

\[\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}\]

\[\Rightarrow f'\left( 0 \right) = 1\]

\[\Rightarrow f''\left( x \right) = \dfrac{{{d^2}}}{{d{x^2}}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{{ - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\]

\[f''\left( 0 \right) = 0\]

Continue this for several values.

Hence the Maclaurin series of \[{\tan ^{ - 1}}\left( x \right)\] is given by

\[\Rightarrow f(x) = {\tan ^{ - 1}}\left( x \right) = x - \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + \cdot \cdot \cdot \]

In general

\[\Rightarrow {\tan ^{ - 1}}\left( x \right) = \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{2n + 1}}} \]

For the given function replace \[x\] by \[ - \dfrac{1}{2}\] , then the Maclaurin series of \[\Rightarrow{\tan ^{ - 1}}\left( x \right)\] becomes

\[\Rightarrow {\tan ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - \dfrac{1}{2} - \dfrac{{{{\left( { - \dfrac{1}{2}} \right)}^3}}}{3} + \dfrac{{{{\left( { - \dfrac{1}{2}} \right)}^5}}}{5} + \cdot \cdot \cdot \]

\[\Rightarrow {\tan ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - \dfrac{1}{2} + \dfrac{1}{{{2^3} \cdot 3}} - \dfrac{1}{{{2^5} \cdot 5}} + \cdot \cdot \cdot \]

\[\Rightarrow {\tan ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - \dfrac{1}{2} + \dfrac{1}{{8 \cdot 3}} - \dfrac{1}{{32 \cdot 5}} + \cdot \cdot \cdot \]

\[\Rightarrow {\tan ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - \dfrac{1}{2} + \dfrac{1}{{24}} - \dfrac{1}{{160}} + \cdot \cdot \cdot \]

\[{\tan ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - 0.5 + 0.04166 - 0.00625 + \cdot \cdot \cdot \]

\[\Rightarrow {\tan ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - 0.46459...\]

Hence the value of \[{\tan ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\] is \[ - 0.465\] rounded to the third decimal figure.

**So, the correct answer is “- 0.465”.**

**Note**: We can calculate the inverse of a trigonometry ratio by the help of a tale of trigonometry ratios for the standard angles. or we can determine the inverse value of tan we use the maclaurin series expansion formula. Hence we substitute the value of x in the given formula and hence we obtain the solution for the question.

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