Factor over the Complex Numbers tan(arccos(6x))

$\tan (\arccos (6 x))$

Draw a triangle in the plane with vertices $(6 x, \sqrt{1^{2} - {(6 x)}^{2}})$ , $(6 x, 0)$ , and the origin. Then $\arccos (6 x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(6 x, \sqrt{1^{2} - {(6 x)}^{2}})$ . Therefore, $\tan (\arccos (6 x))$ is $\frac{\sqrt{1^{2} - {(6 x)}^{2}}}{6 x}$ .

$\frac{\sqrt{1^{2} - {(6 x)}^{2}}}{6 x}$

Simplify the numerator.

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Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - i^{2} = (a + i) (a - i)$ where $a = 1$ and $i = 6 x$ .

$\frac{\sqrt{(1 + 6 x) (1 - (6 x))}}{6 x}$

Multiply $6$ by $- 1$ .

$\frac{\sqrt{(1 + 6 x) (1 - 6 x)}}{6 x}$

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