Find Trig Functions Using Identities csc(theta)=4 , cot(theta)<0

$\csc (θ) = 4$ , $\cot (θ) < 0$

The cotangent function is negative in the second and fourth quadrants. The cosecant function is positive in the first and second quadrants. The set of solutions for $θ$ are limited to the second quadrant since that is the only quadrant found in both sets.

Solution is in the second quadrant.

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

$\csc (θ) = \frac{hypotenuse}{opposite}$

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

$Adjacent = - \sqrt{{hypotenuse}^{2} - {opposite}^{2}}$

Replace the known values in the equation.

$Adjacent = - \sqrt{{(4)}^{2} - {(1)}^{2}}$

Simplify inside the radical.

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Negate $\sqrt{{(4)}^{2} - {(1)}^{2}}$ .

Adjacent $= - \sqrt{{(4)}^{2} - {(1)}^{2}}$

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

Adjacent $= - \sqrt{16 - {(1)}^{2}}$

One to any power is one.

Adjacent $= - \sqrt{16 - 1 \cdot 1}$

Multiply $- 1$ by $1$ .

Adjacent $= - \sqrt{16 - 1}$

Subtract $1$ from $16$ .

Adjacent $= - \sqrt{15}$

Find the value of sine.

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

$\sin (θ) = \frac{opp}{hyp}$

Substitute in the known values.

$\sin (θ) = \frac{1}{4}$

Find the value of cosine.

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

$\cos (θ) = \frac{adj}{hyp}$

Substitute in the known values.

$\cos (θ) = \frac{- \sqrt{15}}{4}$

Move the negative in front of the fraction.

$\cos (θ) = - \frac{\sqrt{15}}{4}$

Find the value of tangent.

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Use the definition of tangent to find the value of $\tan (θ)$ .

$\tan (θ) = \frac{opp}{adj}$

Substitute in the known values.

$\tan (θ) = \frac{1}{- \sqrt{15}}$

Simplify the value of $\tan (θ)$ .

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Cancel the common factor of $1$ and $- 1$ .

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Rewrite $1$ as $- 1 (- 1)$ .

$\tan (θ) = \frac{- 1 \cdot - 1}{- \sqrt{15}}$

Move the negative in front of the fraction.

$\tan (θ) = - \frac{1}{\sqrt{15}}$

Multiply $\frac{1}{\sqrt{15}}$ by $\frac{\sqrt{15}}{\sqrt{15}}$ .

$\tan (θ) = - (\frac{1}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}})$

Combine and simplify the denominator.

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Multiply $\frac{1}{\sqrt{15}}$ and $\frac{\sqrt{15}}{\sqrt{15}}$ .

$\tan (θ) = - \frac{\sqrt{15}}{\sqrt{15} \sqrt{15}}$

Raise $\sqrt{15}$ to the power of $1$ .

$\tan (θ) = - \frac{\sqrt{15}}{\sqrt{15} \sqrt{15}}$

Raise $\sqrt{15}$ to the power of $1$ .

$\tan (θ) = - \frac{\sqrt{15}}{\sqrt{15} \sqrt{15}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\tan (θ) = - \frac{\sqrt{15}}{{\sqrt{15}}^{1 + 1}}$

Add $1$ and $1$ .

$\tan (θ) = - \frac{\sqrt{15}}{{\sqrt{15}}^{2}}$

Rewrite ${\sqrt{15}}^{2}$ as $15$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{15}$ as $15^{\frac{1}{2}}$ .

$\tan (θ) = - \frac{\sqrt{15}}{{(15^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\tan (θ) = - \frac{\sqrt{15}}{15^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\tan (θ) = - \frac{\sqrt{15}}{15^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\tan (θ) = - \frac{\sqrt{15}}{15^{\frac{2}{2}}}$

Rewrite the expression.

$\tan (θ) = - \frac{\sqrt{15}}{15}$

Evaluate the exponent.

$\tan (θ) = - \frac{\sqrt{15}}{15}$

Find the value of cotangent.

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

$\cot (θ) = \frac{adj}{opp}$

Substitute in the known values.

$\cot (θ) = \frac{- \sqrt{15}}{1}$

Divide $- \sqrt{15}$ by $1$ .

$\cot (θ) = - \sqrt{15}$

Find the value of secant.

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

$\sec (θ) = \frac{hyp}{adj}$

Substitute in the known values.

$\sec (θ) = \frac{4}{- \sqrt{15}}$

Simplify the value of $\sec (θ)$ .

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Move the negative in front of the fraction.

$\sec (θ) = - \frac{4}{\sqrt{15}}$

Multiply $\frac{4}{\sqrt{15}}$ by $\frac{\sqrt{15}}{\sqrt{15}}$ .

$\sec (θ) = - (\frac{4}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}})$

Combine and simplify the denominator.

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Multiply $\frac{4}{\sqrt{15}}$ and $\frac{\sqrt{15}}{\sqrt{15}}$ .

$\sec (θ) = - \frac{4 \sqrt{15}}{\sqrt{15} \sqrt{15}}$

Raise $\sqrt{15}$ to the power of $1$ .

$\sec (θ) = - \frac{4 \sqrt{15}}{\sqrt{15} \sqrt{15}}$

Raise $\sqrt{15}$ to the power of $1$ .

$\sec (θ) = - \frac{4 \sqrt{15}}{\sqrt{15} \sqrt{15}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\sec (θ) = - \frac{4 \sqrt{15}}{{\sqrt{15}}^{1 + 1}}$

Add $1$ and $1$ .

$\sec (θ) = - \frac{4 \sqrt{15}}{{\sqrt{15}}^{2}}$

Rewrite ${\sqrt{15}}^{2}$ as $15$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{15}$ as $15^{\frac{1}{2}}$ .

$\sec (θ) = - \frac{4 \sqrt{15}}{{(15^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sec (θ) = - \frac{4 \sqrt{15}}{15^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\sec (θ) = - \frac{4 \sqrt{15}}{15^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\sec (θ) = - \frac{4 \sqrt{15}}{15^{\frac{2}{2}}}$

Rewrite the expression.

$\sec (θ) = - \frac{4 \sqrt{15}}{15}$

Evaluate the exponent.

$\sec (θ) = - \frac{4 \sqrt{15}}{15}$

This is the solution to each trig value.

$\sin (θ) = \frac{1}{4}$

$\cos (θ) = - \frac{\sqrt{15}}{4}$

$\tan (θ) = - \frac{\sqrt{15}}{15}$

$\cot (θ) = - \sqrt{15}$

$\sec (θ) = - \frac{4 \sqrt{15}}{15}$

$\csc (θ) = 4$

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