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**Hint:**We will prove the question by recalling the fact that $\arctan \left( x \right) = {\tan ^{ - 1}}\left( x \right)$. We will use the addition of inverse tangent formula to calculate the value to proceed through the problem. We can make use of the fact that if $\tan \theta = a$ for $a \in R$, then, the value of $\theta $ lies in the interval $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.

**Complete Step by Step Solution:**

From the question, we know that we have to find the value of \[\arctan \left( {\dfrac{1}{2}} \right) + \arctan \left( {\dfrac{1}{3}} \right)\].

Now, we also know that, $\arctan \left( x \right) = {\tan ^{ - 1}}\left( x \right)$

So, \[\arctan \left( {\dfrac{1}{2}} \right) + \arctan \left( {\dfrac{1}{3}} \right)\] can also be written as –

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$

So, the question can also be written as –

${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{4}$

Taking left – hand side from the above equation, we get –

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$

We have to prove the above equation to $\dfrac{\pi }{4}$ . Therefore, we know that, the formula for addition of inverse of tangents is –

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) \cdots \left( 1 \right)$ , if $xy < 1$

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) \cdots \left( 2 \right)$ , if $x > 0,y > 0,xy > 1$

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = - \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) \cdots \left( 3 \right)$ , if $x < 0,y < 0,xy > 1$

Here, $x = \dfrac{1}{2}$ and $y = \dfrac{1}{3}$

$xy = \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$

$\therefore xy > 0$ and $xy < 1$

Therefore, from the above we can conclude that, the values of $x$ and $y$ are greater than 0 but when multiplied with each other gives the value less than 1.

So, now, we will use the equation (1) as it is used when the value of $xy$ is less than 1.

Therefore putting $x = \dfrac{1}{2}$ and $y = \dfrac{1}{3}$ in the equation (1), we get –

$

\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{2} + \dfrac{1}{3}}}{{1 - \dfrac{1}{2} \times \dfrac{1}{3}}}} \right) \\

\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{5}{6}}}{{1 - \dfrac{1}{6}}}} \right) \\

$

Now, solving the denominator in the right – hand side of the above equation, we get –

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{5}{6}}}{{\dfrac{5}{6}}}} \right)$

In the above equation, we have the numerator and denominator same in the right – hand side term, therefore, cancelling them, we get –

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left( 1 \right)$

Now, we know that the value of $\tan $ is 1 when the angle is $\dfrac{\pi }{4}$ . Therefore, ${\tan ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{4}$ , we get –

$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{4}$

**Hence, the value of ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$ is equal to $\dfrac{\pi }{4}$ as it was the required answer to the question.**

**Note:**

Whenever we get these types of problems, we should first of all check whether we need the principal solution or the general solution for the question to be solved. We should only report the angle that was present in the principal range of the inverse of the tangent function.

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