How do you $\arctan \left( {{a}^{2}} \right)-\arctan a-0.22=0$?

Hint: The arctangent function is the same as the inverse tangent function. So we can apply the property of the tangent inverse function which is given by $\arctan x-\arctan y=\arctan \left( \dfrac{x-y}{1+xy} \right)$ on the LHS of the given equation. On applying this identity and after simplifying the equation, we will obtain a cubic equation in terms of a, which can be solved using a graphing calculator.

Complete step by step solution:
The equation given in the above question is written as
$\Rightarrow \arctan \left( {{a}^{2}} \right)-\arctan a-0.22=0$
Adding $0.22$ on both the sides of the above equation, we get
\[\begin{align}
  & \Rightarrow \arctan \left( {{a}^{2}} \right)-\arctan a-0.22+0.22=0+0.22 \\
 & \Rightarrow \arctan \left( {{a}^{2}} \right)-\arctan a=0.22......\left( i \right) \\
\end{align}\]
Now, we know the property of the inverse tangent function which is given by
$\Rightarrow \arctan x-\arctan y=\arctan \left( \dfrac{x-y}{1+xy} \right)$
On putting $x={{a}^{2}}$ and $y=a$ in the above identity, we get
$\begin{align}
  & \Rightarrow \arctan {{a}^{2}}-\arctan a=\arctan \left( \dfrac{{{a}^{2}}-a}{1+{{a}^{2}}\left( a \right)} \right) \\
 & \Rightarrow \arctan {{a}^{2}}-\arctan a=\arctan \left( \dfrac{{{a}^{2}}-a}{1+{{a}^{3}}} \right) \\
\end{align}$
On substituting the above equation in the equation (i) we get
\[\Rightarrow \arctan \left( \dfrac{{{a}^{2}}-a}{1+{{a}^{3}}} \right)=0.22\]
Now, taking tangent on both the sides of the above equation we get
\[\Rightarrow \dfrac{{{a}^{2}}-a}{1+{{a}^{3}}}=\tan \left( 0.22 \right)\]
On putting $\tan \left( 0.22 \right)=0.2236$ we get
\[\Rightarrow \dfrac{{{a}^{2}}-a}{1+{{a}^{3}}}=0.2236\]
Now, on multiplying by \[1+{{a}^{3}}\] on both the sides, we get
\[\begin{align}
  & \Rightarrow {{a}^{2}}-a=0.2236\left( 1+{{a}^{3}} \right) \\
 & \Rightarrow {{a}^{2}}-a=0.2236+0.2236{{a}^{3}} \\
\end{align}\]
On subtracting \[0.2236{{a}^{3}}\] from both the sides, we get
\[\begin{align}
  & \Rightarrow {{a}^{2}}-a-0.2236{{a}^{3}}=0.2236+0.2236{{a}^{3}}-0.2236 \\
 & \Rightarrow {{a}^{2}}-a-0.2236{{a}^{3}}=0.2236 \\
\end{align}\]
Now, multiplying the above equation by $-1$ we get
\[\Rightarrow 0.2236{{a}^{3}}-{{a}^{2}}+a=-0.2236\]
Now, adding \[0.2236\] both the sides, we get
\[\begin{align}
  & \Rightarrow 0.2236{{a}^{3}}-{{a}^{2}}+a+0.2236=-0.2236+0.2236 \\
 & \Rightarrow 0.2236{{a}^{3}}-{{a}^{2}}+a+0.2236=0 \\
\end{align}\]
On solving the above equation using the graphing calculator, we get
$\Rightarrow a=-0.18712,a=2.04100,a=2.61839$
Hence, we have obtained the required solutions of the given equation.

Note: We must note that the equation given in the question is too complicated to be solved using the concepts of elementary mathematics. So we have used the graphing calculator to solve the given equation. We can also use different numerical techniques in order to solve these types of equations.