Find the Cosecant Given the Point (- square root of 3,1)

$(- \sqrt{3}, 1)$

To find the $\csc (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(- \sqrt{3}, 1)$ , draw the triangle between the three points $(0, 0)$ , $(- \sqrt{3}, 0)$ , and $(- \sqrt{3}, 1)$ .

Opposite : $1$

Adjacent : $- \sqrt{3}$

Find the hypotenuse using Pythagorean theorem $c = \sqrt{a^{2} + b^{2}}$ .

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Simplify the expression.

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

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

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

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

Multiply ${\sqrt{3}}^{2}$ by $1$ .

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

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

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

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

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Simplify the expression.

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One to any power is one.

$\sqrt{3 + 1}$

Add $3$ and $1$ .

$\sqrt{4}$

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

$\sqrt{2^{2}}$

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

$2$

$\csc (θ) = \frac{Hypotenuse}{Opposite}$ therefore $\csc (θ) = \frac{2}{1}$ .

$\frac{2}{1}$

Divide $2$ by $1$ .

$\csc (θ) = 2$

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