I think u know elementry trigo

$\tan \theta=x-\frac{1}{4x}$ $\color{#D61F06}{1+\tan^{2}\theta=\sec^{2}\theta}$ $\color{#D61F06}{\Rightarrow}\color{#333333} 1+(x-\frac{1}{4x})^{2}=\sec^{2}\theta$ $\color{#D61F06}{\Rightarrow}\color{#333333}1+x^{2}-\frac{1}{2}+\frac{1}{16x^{2}}=\sec^{2}\theta\color{#D61F06}{\Rightarrow}\color{#333333}x^{2}+\frac{1}{2}+\frac{1}{16x^{2}}=\sec^{2}\theta$ $\color{#D61F06}{\Rightarrow}\color{#333333} (x+\frac{1}{4x})^{2}=\sec^{2}\theta\color{#D61F06}{\Rightarrow}\color{#333333} \sec \theta=\pm(x+\frac{1}{4x})$ $1)\space\space\sec \theta-\tan \theta=(x+\frac{1}{4x})-(x-\frac{1}{4x})=\color{#3D99F6}{\frac{1}{2x}}$ $2)\space\space\sec \theta-\tan \theta=-(x+\frac{1}{4x})-(x-\frac{1}{4x})=\color{#3D99F6}{-2x}\color{#333333}\space\space\space\Box$

First we'll combine into one fraction.

$tan \theta = \frac{4x^2-1}{4x}$

$tan \theta = \frac{opposite}{adjacent}$

We use the above triangle and the Pythagorean Theorem to solve for y.

$(4x^2 - 1)^2 + (4x)^2 = y^2$

$16x^4 +8x^2 +1 = y^2$

$(4x^2 +1)^2 = y^2$

$y= \pm (4x^2 +1)$

We know $sec \theta = \frac{hypotenuse}{adjacent}$

$sec \theta = \frac{y}{4x} = \frac{\pm(4x^2+1)}{4x}$

Thus (plus version),

$sec \theta - \tan \theta = \frac{4x^2+1}{4x} - \frac{4x^2-1}{4x} = \frac{1}{2x}$

OR (minus version)

$sec \theta - \tan \theta = \frac{-(4x^2+1)}{4x} - \frac{4x^2-1}{4x} = -2x$