Precalculus Examples

$\sec (\arctan (\frac{7}{8}))$

Step 1

Draw a triangle in the plane with vertices $(1, \frac{7}{8})$ , $(1, 0)$ , and the origin. Then $\arctan (\frac{7}{8})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, \frac{7}{8})$ . Therefore, $\sec (\arctan (\frac{7}{8}))$ is $\sqrt{1 + {(\frac{7}{8})}^{2}}$ .

$\sqrt{1 + {(\frac{7}{8})}^{2}}$

Step 2

Apply the product rule to $\frac{7}{8}$ .

$\sqrt{1 + \frac{7^{2}}{8^{2}}}$

Step 3

Raise $7$ to the power of $2$ .

$\sqrt{1 + \frac{49}{8^{2}}}$

Step 4

Raise $8$ to the power of $2$ .

$\sqrt{1 + \frac{49}{64}}$

Step 5

Write $1$ as a fraction with a common denominator.

$\sqrt{\frac{64}{64} + \frac{49}{64}}$

Step 6

Combine the numerators over the common denominator.

$\sqrt{\frac{64 + 49}{64}}$

Step 7

Add $64$ and $49$ .

$\sqrt{\frac{113}{64}}$

Step 8

Rewrite $\sqrt{\frac{113}{64}}$ as $\frac{\sqrt{113}}{\sqrt{64}}$ .

$\frac{\sqrt{113}}{\sqrt{64}}$

Step 9

Simplify the denominator.

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Step 9.1

Rewrite $64$ as $8^{2}$ .

$\frac{\sqrt{113}}{\sqrt{8^{2}}}$

Step 9.2

Pull terms out from under the radical, assuming positive real numbers.

$\frac{\sqrt{113}}{8}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{113}}{8}$

Decimal Form:

$1.32876822 \dots$