Find the Exact Value tan(arccos(5x))

$\tan (\arccos (5 x))$

Step 1

Draw a triangle in the plane with vertices $(5 x, \sqrt{1^{2} - {(5 x)}^{2}})$ , $(5 x, 0)$ , and the origin. Then $\arccos (5 x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(5 x, \sqrt{1^{2} - {(5 x)}^{2}})$ . Therefore, $\tan (\arccos (5 x))$ is $\frac{\sqrt{1 - {(5 x)}^{2}}}{5 x}$ .

$\frac{\sqrt{1 - {(5 x)}^{2}}}{5 x}$

Step 2

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

$\frac{\sqrt{1^{2} - {(5 x)}^{2}}}{5 x}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 5 x$ .

$\frac{\sqrt{(1 + 5 x) (1 - (5 x))}}{5 x}$

Multiply $5$ by $- 1$ .

$\frac{\sqrt{(1 + 5 x) (1 - 5 x)}}{5 x}$

Find the Exact Value tan(arccos(5x))

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