Given $\sec A = \frac{\sqrt{61}}{6}$ and that angle $A$ is in Quadrant I, find the exact value of $\csc A$ in simplest radical form using a rational denominator.

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Q. Given

\sec A = \frac{\sqrt{61}}{6}

and that angle

A

is in Quadrant I, find the exact value of

\csc A

in simplest radical form using a rational denominator.

Understand relationship sec $A$ cos $A$ : Understand the relationship between sec $A$ and cos $A$ . $\newline$ Secant is the reciprocal of cosine. $\newline$ $\sec A = \frac{1}{\cos A}$ $\newline$ Given $\sec A = \frac{\sqrt{61}}{6}$ , we can find cos $A$ by taking the reciprocal of sec $A$ .
Calculate $\cos A$ : Calculate $\cos A$ .
$\cos A = \frac{1}{\sec A}$
$\cos A = \frac{1}{\left(\frac{\sqrt{61}}{6}\right)}$
$\cos A = \frac{6}{\sqrt{61}}$
To rationalize the denominator, multiply the numerator and denominator by $\sqrt{61}$ .
$\cos A = \left(\frac{6}{\sqrt{61}}\right) \times \left(\frac{\sqrt{61}}{\sqrt{61}}\right)$
$\cos A = \frac{6 \times \sqrt{61}}{61}$
Use Pythagorean identity find sin A: Use the Pythagorean identity to find $\sin A$ . The Pythagorean identity states that $\sin^2 A + \cos^2 A = 1$ . We know $\cos A$ , so we can solve for $\sin^2 A$ . $\sin^2 A = 1 - \cos^2 A$ $\sin^2 A = 1 - \left(\frac{6 \cdot \sqrt{61}}{61}\right)^2$ $\sin^2 A = 1 - \frac{36 \cdot 61}{61 \cdot 61}$ $\sin^2 A = 1 - \frac{36 \cdot 61}{3721}$ $\sin^2 A = \frac{3721 - 2196}{3721}$ $\sin^2 A = \frac{1525}{3721}$ Since A is in Quadrant I, $\sin A$ is positive. $\sin^2 A + \cos^2 A = 1$ $1$ $\sin^2 A + \cos^2 A = 1$ $2$
Find $\csc A$ : Find $\csc A$ , which is the reciprocal of $\sin A$ .
$\csc A = \frac{1}{\sin A}$
$\csc A = \frac{1}{(\sqrt{1525} / 61)}$
To rationalize the denominator, multiply the numerator and denominator by $\sqrt{1525}$ .
$\csc A = (\frac{61}{\sqrt{1525}}) \cdot (\frac{\sqrt{1525}}{\sqrt{1525}})$
$\csc A = (\frac{61 \cdot \sqrt{1525}}{1525})$
Simplify expression $\csc A$ : Simplify the expression for $\csc A$ . We need to simplify $\sqrt{1525}$ . Since $1525 = 25 \times 61$ , we can take the square root of $25$ out of the radical. $\csc A = \frac{61 \times \sqrt{25 \times 61}}{1525}$ $\csc A = \frac{61 \times 5 \times \sqrt{61}}{1525}$ $\csc A = \frac{305 \times \sqrt{61}}{1525}$ Now, simplify the fraction by dividing both the numerator and the denominator by $305$ . $\csc A = \frac{\sqrt{61}}{5}$