${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\tan }{x}dx$ is equal to

The correct option is:
$\frac{\pi }{12}$

Step-by-step Explanation:

Let the given definite integral be denoted by $I$:
$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\tan x}{\tan x+\cot x}dx$$
Let this be equation (1).

We can solve this definite integral using the standard integration property:
$${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$$

Here, the lower limit is $a=\frac{\pi }{6}$ and the upper limit is $b=\frac{\pi }{3}$.
Calculating the sum of the limits:
$$a+b=\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi +2\pi }{6}=\frac{3\pi }{6}=\frac{\pi }{2}$$

Applying the integration property by replacing $x$ with $\left(\frac{\pi }{2}-x\right)$ in equation (1):
$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\tan \left(\frac{\pi }{2},-,x\right)}{\tan \left(\frac{\pi }{2},-,x\right)}dx$$

Using the complementary angle trigonometric identities $\tan \left(\frac{\pi }{2}-x\right)=\cot x$ and $\cot \left(\frac{\pi }{2}-x\right)=\tan x$, we get:
$$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\cot x}{\cot x+\tan x}dx$$
Let this be equation (2).

Now, add equation (1) and equation (2):
$$I+I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\tan x}{\tan x+\cot x}dx+{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\cot x}{\tan x+\cot x}dx$$

Combining the integrands:
$$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\tan x+\cot x}{\tan x+\cot x}dx$$

Simplifying the term inside the integral:
$$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}1dx$$

Integrating $1$ with respect to $x$:
$$2I={\left[x\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}$$

Evaluating the limits:
$$2I=\frac{\pi }{3}-\frac{\pi }{6}$$
$$2I=\frac{\pi }{6}$$

Divide both sides by $2$ to find $I$:
$$I=\frac{\pi }{12}$$