The problem is to simplify the expression $\tan(x) \cot(9^\circ) + \tan(5x)$. I assume the question is asking to simplify the expression and there is some equality sign hidden. Based on the identity for tangent of a sum, the problem seems to be leading to finding a value of $x$.

1. Problem Description

The problem is to simplify the expression $\tan(x) \cot(9^\circ) + \tan(5x)$. I assume the question is asking to simplify the expression and there is some equality sign hidden. Based on the identity for tangent of a sum, the problem seems to be leading to finding a value of $x$.

2. Solution Steps

Let's consider the tangent addition formula:

$\tan(a+b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$

We have the expression $\tan(x) \cot(9^\circ) + \tan(5x)$. Since $\cot(9^\circ) = \frac{1}{\tan(9^\circ)}$, we can rewrite the expression as:

$\frac{\tan(x)}{\tan(9^\circ)} + \tan(5x)$.

If the original problem asks to simplify the expression $\frac{\tan(x)}{\tan(9^\circ)} + \tan(5x)$, it is not clear what could be simplified further. Let us try to consider a scenario where this is equal to a tangent of a sum:

Let's suppose $\frac{\tan(x)}{\tan(9^\circ)} + \tan(5x) = \tan(a+b)$

Assume we want to get an identity of the form $\tan(A) = \tan(x) \cot(9^\circ) + \tan(5x)$ for some $A$.

Consider the tangent addition formula $\tan(a+b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$

We rewrite the expression as:

$\tan(x)\cot(9^{\circ}) + \tan(5x) = \frac{\tan(x)}{\tan(9^{\circ})} + \tan(5x)$

Multiplying by $\tan(9^{\circ})$, we get $\tan(x) + \tan(5x)\tan(9^{\circ})$

If we set $x = 9^\circ$, then the expression becomes $\tan(9^\circ) \cot(9^\circ) + \tan(5\cdot 9^\circ) = 1 + \tan(45^\circ) = 1 + 1 = 2$

Suppose the original question asked to evaluate the expression when x = $9^\circ$.

Then, $\tan(9^\circ) \cot(9^\circ) + \tan(5\cdot 9^\circ) = \tan(9^\circ) \frac{1}{\tan(9^\circ)} + \tan(45^\circ) = 1 + 1 = 2$.

Let's assume we are supposed to find when $\tan(x) \cot(9^\circ) + \tan(5x) = \tan(6x+9^\circ)$.

Using $\tan(6x+9^\circ) = \frac{\tan(6x) + \tan(9^\circ)}{1 - \tan(6x)\tan(9^\circ)} = \tan(x) \cot(9^\circ) + \tan(5x) = \frac{\tan(x)}{\tan(9^\circ)} + \tan(5x)$,

Then we would have

$\frac{\tan(6x) + \tan(9^\circ)}{1 - \tan(6x)\tan(9^\circ)} = \frac{\tan(x) + \tan(5x)\tan(9^\circ)}{\tan(9^\circ)}$

$\tan(9^\circ)(\tan(6x)+\tan(9^\circ)) = (1 - \tan(6x)\tan(9^\circ))(\tan(x) + \tan(5x)\tan(9^\circ))$

$\tan(9^\circ)\tan(6x) + \tan^2(9^\circ) = \tan(x) + \tan(5x)\tan(9^\circ) - \tan(x)\tan(6x)\tan(9^\circ) - \tan(5x)\tan(6x)\tan^2(9^\circ)$

This becomes complex.

3. Final Answer

Without more context or constraints, it's impossible to determine a simplified answer. If $x=9^\circ$, then the expression equals

2.