If csc(θ) = 17/8 and 0° < θ < 90°, what is tan(θ)? Write your answer in simplified, rationalized form. tan(θ) = ______

Find sin(θ): Use the Pythagorean identity for sine and cosine: sin^2(θ) + cos^2(θ) = 1. Since csc(θ) is the reciprocal of sin(θ), we can find sin(θ) first. sin(θ) = 1 / csc(θ) = 1 / (17/8) = 8/17.Find cos(θ): Now, use the Pythagorean identity to find cos(θ): cos^2(θ) = 1 - sin^2(θ) = 1 - (8/17)^2. Calculate cos^2(θ) = 1 - (64/289).Simplify cos^2(θ): Simplify the expression for cos^2(θ): cos^2(θ) = (289/289) - (64/289) = (225/289). Take the square root to find cos(θ), remembering that since 0° < θ < 90°, cos(θ) will be positive. cos(θ) = √(225/289) = 15/17.Find cos(θ): Now, use the definition of tangent, which is tan(θ) = sin(θ) / cos(θ). Substitute the values of sin(θ) and cos(θ) into the equation: tan(θ) = (8/17) / (15/17).Find tan(θ): Simplify the expression for tan(θ): tan(θ) = (8/17) * (17/15). The 17s cancel out, leaving tan(θ) = 8/15.

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If $\csc(\theta) = \frac{17}{8}$ and 0^\circ < \theta < 90^\circ , what is $\tan(\theta)$ ? $\newline$ Write your answer in simplified, rationalized form. $\newline$ $\tan(\theta) =$ ______ $\newline$

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Algebra 2

Find trigonometric ratios using multiple identities

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Q. If

\csc(\theta) = \frac{17}{8}

and

0^\circ < \theta < 90^\circ

, what is

\tan(\theta)

\newline

Write your answer in simplified, rationalized form.

\newline

\tan(\theta) =

______

\newline

Find $\sin(\theta)$ : Use the Pythagorean identity for sine and cosine: $\sin^2(\theta) + \cos^2(\theta) = 1$ . Since $\csc(\theta)$ is the reciprocal of $\sin(\theta)$ , we can find $\sin(\theta)$ first. $\newline$ $\sin(\theta) = \frac{1}{\csc(\theta)} = \frac{1}{\left(\frac{17}{8}\right)} = \frac{8}{17}$ .
Find $\cos(\theta):$ Now, use the Pythagorean identity to find $\cos(\theta): \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{8}{17}\right)^2.$ $\newline$ Calculate $\cos^2(\theta) = 1 - \left(\frac{64}{289}\right).$
Simplify $\cos^2(\theta)$ : Simplify the expression for $\cos^2(\theta)$ : $\cos^2(\theta) = \frac{289}{289} - \frac{64}{289} = \frac{225}{289}$ . $\newline$ Take the square root to find $\cos(\theta)$ , remembering that since 0^\circ < \theta < 90^\circ, $\cos(\theta)$ will be positive. $\newline$ $\cos(\theta) = \sqrt{\frac{225}{289}} = \frac{15}{17}$ .
Find $\cos(\theta)$ : Now, use the definition of tangent, which is $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ . $\newline$ Substitute the values of $\sin(\theta)$ and $\cos(\theta)$ into the equation: $\tan(\theta) = \frac{\frac{8}{17}}{\frac{15}{17}}$ .
Find $\tan(\theta)$ : Simplify the expression for $\tan(\theta)$ : $\tan(\theta) = \frac{8}{17} \cdot \frac{17}{15}$ . $\newline$ The $17$ s cancel out, leaving $\tan(\theta) = \frac{8}{15}$ .