Let's solve the given mathematical expression step-by-step.
We start with the given expression:
[tex]$
\frac{\csc(6x) + \cot(6x)}{\csc(6x) - \cot(6x)}
$[/tex]
First, recall the definitions of cosecant ([tex]\(\csc\)[/tex]) and cotangent ([tex]\(\cot\)[/tex]) in terms of sine ([tex]\(\sin\)[/tex]) and cosine ([tex]\(\cos\)[/tex]):
[tex]$
\csc(6x) = \frac{1}{\sin(6x)}
$[/tex]
[tex]$
\cot(6x) = \frac{\cos(6x)}{\sin(6x)}
$[/tex]
Using these definitions, we can rewrite the numerator and the denominator of the given expression.
The numerator [tex]\(\csc(6x) + \cot(6x)\)[/tex] becomes:
[tex]$
\csc(6x) + \cot(6x) = \frac{1}{\sin(6x)} + \frac{\cos(6x)}{\sin(6x)} = \frac{1 + \cos(6x)}{\sin(6x)}
$[/tex]
Similarly, the denominator [tex]\(\csc(6x) - \cot(6x)\)[/tex] becomes:
[tex]$
\csc(6x) - \cot(6x) = \frac{1}{\sin(6x)} - \frac{\cos(6x)}{\sin(6x)} = \frac{1 - \cos(6x)}{\sin(6x)}
$[/tex]
Now, substituting these back into our original expression, we get:
[tex]$
\frac{\frac{1 + \cos(6x)}{\sin(6x)}}{\frac{1 - \cos(6x)}{\sin(6x)}}
$[/tex]
We can simplify this fraction by multiplying the numerator by the reciprocal of the denominator:
[tex]$
\frac{(1 + \cos(6x)) / \sin(6x)}{(1 - \cos(6x)) / \sin(6x)} = \frac{1 + \cos(6x)}{1 - \cos(6x)}
$[/tex]
Next, we use the trigonometric identities:
[tex]$
1 + \cos(6x) = 2 \cos^2(3x) \quad \text{(since } \cos(6x) = 2 \cos^2(3x) - 1 \text{)}
$[/tex]
[tex]$
1 - \cos(6x) = 2 \sin^2(3x) \quad \text{(since } \cos(6x) = 1 - 2 \sin^2(3x) \text{)}
$[/tex]
Hence, the expression simplifies to:
[tex]$
\frac{2 \cos^2(3x)}{2 \sin^2(3x)} = \frac{\cos^2(3x)}{\sin^2(3x)} = \cot^2(3x)
$[/tex]
Thus, the given expression simplifies to [tex]\(\cot^2(3x)\)[/tex].
Therefore, the correct answer is:
[tex]$
\boxed{B.\ \cot^2(3x)}
$[/tex]
