Given $\tan A=-\frac{11}{60}$ and that angle $A$ is in Quadrant IV, find the exact value of $\csc A$ in simplest radical form using a rational denominator.

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Q. Given

\tan A=-\frac{11}{60}

and that angle

A

is in Quadrant IV, find the exact value of

\csc A

in simplest radical form using a rational denominator.

Given Information: We are given that $\tan A = -\frac{11}{60}$ and that angle $A$ is in Quadrant IV. In the fourth quadrant, the tangent function is negative because sine is negative and cosine is positive. We can use the Pythagorean identity for tangent, which is $\tan^2 A + 1 = \sec^2 A$ , to find $\sec A$ and then use that to find $\csc A$ .
Find sec A: First, let's find sec A using the identity $\tan^2 A + 1 = \sec^2 A$ . We substitute $\tan A$ with $-(11/60)$ to find $\sec^2 A$ . $\newline$ $\sec^2 A = \tan^2 A + 1$ $\newline$ $\sec^2 A = (-11/60)^2 + 1$ $\newline$ $\sec^2 A = (121/3600) + 1$ $\newline$ $\sec^2 A = (121/3600) + (3600/3600)$ $\newline$ $\sec^2 A = (121 + 3600) / 3600$ $\newline$ $\sec^2 A = 3721 / 3600$
Find $\csc A$ : Now, we take the square root of $\sec^2 A$ to find $\sec A$ . Since $A$ is in the fourth quadrant, $\sec A$ will be positive. $\newline$ $\sec A = \sqrt{\frac{3721}{3600}}$ $\newline$ $\sec A = \frac{\sqrt{3721}}{\sqrt{3600}}$ $\newline$ $\sec A = \frac{61}{60}$
Find $\cos A$ : The cosecant function is the reciprocal of the sine function, so $\csc A = \frac{1}{\sin A}$ . Since $\sec A$ is the reciprocal of $\cos A$ , and we have the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ , we can find $\sin A$ using the relationship $\sin^2 A = 1 - \cos^2 A$ .
Find $\sin A$ : We know that $\cos A = \frac{1}{\sec A}$ , so we can find $\cos A$ first. $\newline$ $\cos A = \frac{1}{\sec A}$ $\newline$ $\cos A = \frac{1}{\frac{61}{60}}$ $\newline$ $\cos A = \frac{60}{61}$
Find $\csc A$ : Now we use the identity $\sin^2 A + \cos^2 A = 1$ to find $\sin^2 A$ .
$\sin^2 A = 1 - \cos^2 A$
$\sin^2 A = 1 - \left(\frac{60}{61}\right)^2$
$\sin^2 A = 1 - \left(\frac{3600}{3721}\right)$
$\sin^2 A = \left(\frac{3721}{3721}\right) - \left(\frac{3600}{3721}\right)$
$\sin^2 A = \frac{3721 - 3600}{3721}$
$\sin^2 A = \frac{121}{3721}$
Find $\csc A$ : Now we use the identity $\sin^2 A + \cos^2 A = 1$ to find $\sin^2 A$ .
$\sin^2 A = 1 - \cos^2 A$
$\sin^2 A = 1 - \left(\frac{60}{61}\right)^2$
$\sin^2 A = 1 - \left(\frac{3600}{3721}\right)$
$\sin^2 A = \left(\frac{3721}{3721}\right) - \left(\frac{3600}{3721}\right)$
$\sin^2 A = \frac{3721 - 3600}{3721}$
$\sin^2 A = \frac{121}{3721}$ We take the square root of $\sin^2 A$ to find $\sin^2 A + \cos^2 A = 1$ $0$ . Since $\sin^2 A + \cos^2 A = 1$ $1$ is in the fourth quadrant, $\sin^2 A + \cos^2 A = 1$ $0$ will be negative.
$\sin^2 A + \cos^2 A = 1$ $3$
$\sin^2 A + \cos^2 A = 1$ $4$
$\sin^2 A + \cos^2 A = 1$ $5$
Find $csc A$ : Now we use the identity $\sin^2 A + \cos^2 A = 1$ to find $\sin^2 A$ .
$\sin^2 A = 1 - \cos^2 A$
$\sin^2 A = 1 - \left(\frac{60}{61}\right)^2$
$\sin^2 A = 1 - \left(\frac{3600}{3721}\right)$
$\sin^2 A = \left(\frac{3721}{3721}\right) - \left(\frac{3600}{3721}\right)$
$\sin^2 A = \frac{3721 - 3600}{3721}$
$\sin^2 A = \frac{121}{3721}$ We take the square root of $\sin^2 A$ to find $\sin^2 A + \cos^2 A = 1$ $0$ . Since $\sin^2 A + \cos^2 A = 1$ $1$ is in the fourth quadrant, $\sin^2 A + \cos^2 A = 1$ $0$ will be negative.
$\sin^2 A + \cos^2 A = 1$ $3$
$\sin^2 A + \cos^2 A = 1$ $4$
$\sin^2 A + \cos^2 A = 1$ $5$ Finally, we find $csc A$ , which is the reciprocal of $\sin^2 A + \cos^2 A = 1$ $0$ .
$\sin^2 A + \cos^2 A = 1$ $8$
$\sin^2 A + \cos^2 A = 1$ $9$
$\sin^2 A$ $0$