Precalculus Examples

$\tan (θ) = \frac{12}{5}$

Step 1

Use the definition of tangent to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.

$\tan (θ) = \frac{opposite}{adjacent}$

Step 2

Find the hypotenuse of the unit circle triangle. Since the opposite and adjacent sides are known, use the Pythagorean theorem to find the remaining side.

$Hypotenuse = \sqrt{{opposite}^{2} + {adjacent}^{2}}$

Step 3

Replace the known values in the equation.

$Hypotenuse = \sqrt{{(12)}^{2} + {(5)}^{2}}$

Step 4

Simplify inside the radical.

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Step 4.1

Raise $12$ to the power of $2$ .

Hypotenuse $= \sqrt{144 + {(5)}^{2}}$

Step 4.2

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

Hypotenuse $= \sqrt{144 + 25}$

Step 4.3

Add $144$ and $25$ .

Hypotenuse $= \sqrt{169}$

Step 4.4

Rewrite $169$ as $13^{2}$ .

Hypotenuse $= \sqrt{13^{2}}$

Step 4.5

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

Hypotenuse $= 13$

Step 5

Find the value of sine.

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Step 5.1

Use the definition of sine to find the value of $\sin (θ)$ .

$\sin (θ) = \frac{opp}{hyp}$

Step 5.2

Substitute in the known values.

$\sin (θ) = \frac{12}{13}$

Step 6

Find the value of cosine.

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Step 6.1

Use the definition of cosine to find the value of $\cos (θ)$ .

$\cos (θ) = \frac{adj}{hyp}$

Step 6.2

Substitute in the known values.

$\cos (θ) = \frac{5}{13}$

Step 7

Find the value of cotangent.

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Step 7.1

Use the definition of cotangent to find the value of $\cot (θ)$ .

$\cot (θ) = \frac{adj}{opp}$

Step 7.2

Substitute in the known values.

$\cot (θ) = \frac{5}{12}$

Step 8

Find the value of secant.

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Step 8.1

Use the definition of secant to find the value of $\sec (θ)$ .

$\sec (θ) = \frac{hyp}{adj}$

Step 8.2

Substitute in the known values.

$\sec (θ) = \frac{13}{5}$

Step 9

Find the value of cosecant.

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Step 9.1

Use the definition of cosecant to find the value of $\csc (θ)$ .

$\csc (θ) = \frac{hyp}{opp}$

Step 9.2

Substitute in the known values.

$\csc (θ) = \frac{13}{12}$

Step 10

This is the solution to each trig value.

$\sin (θ) = \frac{12}{13}$

$\cos (θ) = \frac{5}{13}$

$\tan (θ) = \frac{12}{5}$

$\cot (θ) = \frac{5}{12}$

$\sec (θ) = \frac{13}{5}$

$\csc (θ) = \frac{13}{12}$