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If csc⁡(θ)=178 \csc(\theta) = \frac{17}{8} and 0^\circ < \theta < 90^\circ , what is tan⁡(θ) \tan(\theta) ?\newlineWrite your answer in simplified, rationalized form.\newlinetan⁡(θ)= \tan(\theta) = ______\newline

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Q. If csc⁡(θ)=178 \csc(\theta) = \frac{17}{8} and 0∘<θ<90∘ 0^\circ < \theta < 90^\circ , what is tan⁡(θ) \tan(\theta) ?\newlineWrite your answer in simplified, rationalized form.\newlinetan⁡(θ)= \tan(\theta) = ______\newline
  1. Find sin⁡(θ)\sin(\theta): Use the Pythagorean identity for sine and cosine: sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1. Since csc⁡(θ)\csc(\theta) is the reciprocal of sin⁡(θ)\sin(\theta), we can find sin⁡(θ)\sin(\theta) first.\newlinesin⁡(θ)=1csc⁡(θ)=1(178)=817\sin(\theta) = \frac{1}{\csc(\theta)} = \frac{1}{\left(\frac{17}{8}\right)} = \frac{8}{17}.
  2. Find cos⁡(θ):\cos(\theta): Now, use the Pythagorean identity to find cos⁡(θ):cos⁡2(θ)=1−sin⁡2(θ)=1−(817)2.\cos(\theta): \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{8}{17}\right)^2.\newlineCalculate cos⁡2(θ)=1−(64289).\cos^2(\theta) = 1 - \left(\frac{64}{289}\right).
  3. Simplify cos⁡2(θ)\cos^2(\theta): Simplify the expression for cos⁡2(θ)\cos^2(\theta): cos⁡2(θ)=289289−64289=225289\cos^2(\theta) = \frac{289}{289} - \frac{64}{289} = \frac{225}{289}.\newlineTake the square root to find cos⁡(θ)\cos(\theta), remembering that since 0^\circ < \theta < 90^\circ, cos⁡(θ)\cos(\theta) will be positive.\newlinecos⁡(θ)=225289=1517\cos(\theta) = \sqrt{\frac{225}{289}} = \frac{15}{17}.
  4. Find cos⁡(θ)\cos(\theta): Now, use the definition of tangent, which is tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.\newlineSubstitute the values of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) into the equation: tan⁡(θ)=8171517\tan(\theta) = \frac{\frac{8}{17}}{\frac{15}{17}}.
  5. Find tan⁡(θ)\tan(\theta): Simplify the expression for tan⁡(θ)\tan(\theta): tan⁡(θ)=817⋅1715\tan(\theta) = \frac{8}{17} \cdot \frac{17}{15}.\newlineThe 1717s cancel out, leaving tan⁡(θ)=815\tan(\theta) = \frac{8}{15}.

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