Precalculus Examples

$\sec (\arctan (\frac{- 3}{5}))$

Step 1

Draw a triangle in the plane with vertices $(1, \frac{- 3}{5})$ , $(1, 0)$ , and the origin. Then $\arctan (\frac{- 3}{5})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, \frac{- 3}{5})$ . Therefore, $\sec (\arctan (\frac{- 3}{5}))$ is $\sqrt{1 + {(- \frac{3}{5})}^{2}}$ .

$\sqrt{1 + {(- \frac{3}{5})}^{2}}$

Step 2

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 2.1

Apply the product rule to $- \frac{3}{5}$ .

$\sqrt{1 + {(- 1)}^{2} {(\frac{3}{5})}^{2}}$

Step 2.2

Apply the product rule to $\frac{3}{5}$ .

$\sqrt{1 + {(- 1)}^{2} \frac{3^{2}}{5^{2}}}$

Step 3

Raise $- 1$ to the power of $2$ .

$\sqrt{1 + 1 \frac{3^{2}}{5^{2}}}$

Step 4

Multiply $\frac{3^{2}}{5^{2}}$ by $1$ .

$\sqrt{1 + \frac{3^{2}}{5^{2}}}$

Step 5

Raise $3$ to the power of $2$ .

$\sqrt{1 + \frac{9}{5^{2}}}$

Step 6

Raise $5$ to the power of $2$ .

$\sqrt{1 + \frac{9}{25}}$

Step 7

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

$\sqrt{\frac{25}{25} + \frac{9}{25}}$

Step 8

Combine the numerators over the common denominator.

$\sqrt{\frac{25 + 9}{25}}$

Step 9

Add $25$ and $9$ .

$\sqrt{\frac{34}{25}}$

Step 10

Rewrite $\sqrt{\frac{34}{25}}$ as $\frac{\sqrt{34}}{\sqrt{25}}$ .

$\frac{\sqrt{34}}{\sqrt{25}}$

Step 11

Simplify the denominator.

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Step 11.1

Rewrite $25$ as $5^{2}$ .

$\frac{\sqrt{34}}{\sqrt{5^{2}}}$

Step 11.2

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

$\frac{\sqrt{34}}{5}$

Step 12

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{34}}{5}$

Decimal Form:

$1.16619037 \dots$