Trigonometry Examples

$\sec (\arctan (\frac{3}{4}))$

Step 1

Draw a triangle in the plane with vertices $(1, \frac{3}{4})$ , $(1, 0)$ , and the origin. Then $\arctan (\frac{3}{4})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, \frac{3}{4})$ . Therefore, $\sec (\arctan (\frac{3}{4}))$ is $\sqrt{1 + {(\frac{3}{4})}^{2}}$ .

$\sqrt{1 + {(\frac{3}{4})}^{2}}$

Step 2

Apply the product rule to $\frac{3}{4}$ .

$\sqrt{1 + \frac{3^{2}}{4^{2}}}$

Step 3

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

$\sqrt{1 + \frac{9}{4^{2}}}$

Step 4

Raise $4$ to the power of $2$ .

$\sqrt{1 + \frac{9}{16}}$

Step 5

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

$\sqrt{\frac{16}{16} + \frac{9}{16}}$

Step 6

Combine the numerators over the common denominator.

$\sqrt{\frac{16 + 9}{16}}$

Step 7

Add $16$ and $9$ .

$\sqrt{\frac{25}{16}}$

Step 8

Rewrite $\sqrt{\frac{25}{16}}$ as $\frac{\sqrt{25}}{\sqrt{16}}$ .

$\frac{\sqrt{25}}{\sqrt{16}}$

Step 9

Simplify the numerator.

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Step 9.1

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

$\frac{\sqrt{5^{2}}}{\sqrt{16}}$

Step 9.2

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

$\frac{5}{\sqrt{16}}$

Step 10

Simplify the denominator.

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Step 10.1

Rewrite $16$ as $4^{2}$ .

$\frac{5}{\sqrt{4^{2}}}$

Step 10.2

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

$\frac{5}{4}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$\frac{5}{4}$

Decimal Form:

$1.25$

Mixed Number Form:

$1 \frac{1}{4}$