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**Hint:**For the given question we have to divide denominator and numerator with $\sin x$. It is known that $\dfrac{{\cos x}}{{\sin x}} = \cot x$. Then it is given that $16\cot x = 12$, we will substitute the value of $\cot x$ and after simplifying we the value of the given question.

**Complete step-by-step answer:**Given $16\cot x = 12$simplifying, we get,

$ \Rightarrow \cot x = \dfrac{3}{4}$ ……………. $\left( i \right)$

Now according to question

$\left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right]$

Dividing denominator and numerator with $\sin x$,we get,

$ \Rightarrow \left[ {\dfrac{{1 + \dfrac{{\cos x}}{{\sin x}}}}{{1 - \dfrac{{\cos x}}{{\sin x}}}}} \right]$

Substituting $\dfrac{{\cos x}}{{\sin x}} = \cot x$, we get,

$ \Rightarrow \left[ {\dfrac{{1 + \cot x}}{{1 - \cot x}}} \right]$

Now, substituting $\left( i \right)$and on simplifying we get,

$ \Rightarrow \left[ {\dfrac{{1 + \dfrac{3}{4}}}{{1 - \dfrac{3}{4}}}} \right]$

$ = 7$

$\therefore \left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right] = 7$

**Answer is option (A)**

**Note:**Alternative method

$\left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right]$

Dividing denominator and numerator with$\cos x$we get

$ \Rightarrow \left[ {\dfrac{{\tan x + 1}}{{\tan x - 1}}} \right]$

We know that $\cot x = \dfrac{3}{4}$and $\tan x = \dfrac{1}{{\cot x}}$, we get,

$\therefore \tan x = \dfrac{4}{3}$………….. (ii)

Now substituting (ii) and simplifying we get

$ \Rightarrow \left[ {\dfrac{{\dfrac{4}{3} + 1}}{{\dfrac{4}{3} - 1}}} \right] = 7$

$\therefore \left[ {\dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right] = 7$

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