How do you evaluate \[\arctan \left( { - \dfrac{3}{4}} \right)\] \[?\]

Hint: Here in this question the given function contains arc, so it comes under the topic of inverse trigonometry function. To find the value of this function we use Maclaurin series or expansion, it’s formula can be defined as \[f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( {{x_0}} \right)}}{{n!}}} \left( {x - {x_0}} \right)\] further solve this formula for \[\arctan \] by simplification we get the required result.

Complete step-by-step answer:
The given function is an inverse trigonometry function. The basic inverse trigonometric functions are used to find the angles in right triangles.
To solve the given inverse trigonometry function we use a Maclaurin series or expansion.
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
 \[f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( {{x_0}} \right)}}{{n!}}} {\left( {x - {x_0}} \right)^n}\]
Where \[f'({x_0})\] , \[f''({x_0})\] , \[f'''({x_0})\] , … are the successive differentials when \[{x_0} = 0\] .
Consider the given function \[\arctan \left( { - \dfrac{3}{4}} \right)\]
Before going to solve this, first find the Maclaurin series of \[\arctan \left( x \right)\] when \[{x_0} = 0\] .
So it is given as
 \[f(x) = \arctan (x)\]
 \[ \Rightarrow f\left( 0 \right) = 0\]
 \[f'\left( x \right) = \dfrac{d}{{dx}}\arctan (x) = \dfrac{1}{{1 + {x^2}}}\]
 \[ \Rightarrow f'\left( 0 \right) = 1\]
 \[f''\left( x \right) = \dfrac{{{d^2}}}{{d{x^2}}}\arctan (x) = \dfrac{d}{{dx}}\left( {\dfrac{1}{{1 + {x^2}}}} \right) = \dfrac{{ - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\]
 \[f''\left( 0 \right) = 0\]
And so on.
Continuing like this for other derivative values.
Hence the Maclaurin series of \[\arctan (x)\] is given by
 \[f(x) = \arctan (x) = x - \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + \cdot \cdot \cdot \]
In general it is defined as
 \[\arctan (x) = \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{2n + 1}}} \]
For the given function we have to replace \[x\] by \[ - \dfrac{3}{4}\] , then the Maclaurin series of \[\arctan (x)\] becomes
 \[\arctan \left( { - \dfrac{3}{4}} \right) = - \dfrac{3}{4} - \dfrac{{{{\left( { - \dfrac{3}{4}} \right)}^3}}}{3} + \dfrac{{{{\left( { - \dfrac{3}{4}} \right)}^5}}}{5} + \cdot \cdot \cdot \]
On simplifying we have
 \[ \Rightarrow \arctan \left( { - \dfrac{3}{4}} \right) = - \dfrac{3}{4} + \dfrac{{{3^3}}}{{{4^3} \cdot 3}} - \dfrac{{{3^5}}}{{{4^5} \cdot 5}} + \cdot \cdot \cdot \]
Expand the exponential numbers we have
 \[ \Rightarrow \arctan \left( { - \dfrac{3}{4}} \right) = - \dfrac{3}{4} + \dfrac{{27}}{{64 \cdot 3}} - \dfrac{{243}}{{256 \cdot 5}} + \cdot \cdot \cdot \]
 \[ \Rightarrow \arctan \left( { - \dfrac{3}{4}} \right) = - \dfrac{3}{4} + \dfrac{{27}}{{192}} - \dfrac{{243}}{{1280}} + \cdot \cdot \cdot \]
On further simplification we get
 \[ \Rightarrow \arctan \left( { - \dfrac{3}{4}} \right) = - 0.75 + 0.140625 - 0.18984375 + \cdot \cdot \cdot \]
On adding these numbers we have
 \[ \Rightarrow \arctan \left( { - \dfrac{3}{4}} \right) = - 0.799219\]
Hence the value of \[\arctan \left( { - \dfrac{3}{4}} \right)\] is \[ - 0.780\] rounded to the third decimal figure.
So, the correct answer is “ - 0.780”.

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.