Find $\sin θ$ .
$\sec θ = \frac{13}{6}, \tan θ < 0$
$\sin θ =$
(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Answer

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Final Answer: $\sin θ = - \frac{115}{130}$

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Step 1 ：We know that $\sec θ = \frac{1}{\cos θ}$ , so $\cos θ = \frac{1}{\sec θ} = \frac{6}{13}$ .

Step 2 ：We also know that $\sin^{2} θ + \cos^{2} θ = 1$ , so we can solve for $\sin θ$ by rearranging this equation to get $\sin θ = \sqrt{1 - \cos^{2} θ}$ .

Step 3 ：However, we also know that $\tan θ < 0$ , which means that $\sin θ$ and $\cos θ$ must have opposite signs. Since $\cos θ > 0$ , $\sin θ$ must be negative.

Step 4 ：Therefore, $\sin θ = - \sqrt{1 - \cos^{2} θ}$ .

Step 5 ：Final Answer: $\sin θ = - \frac{115}{130}$

Find sin⁡θ.sec⁡θ=136,tan⁡θ<0sin⁡θ=(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

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Find $\sin θ$ .
$\sec θ = \frac{13}{6}, \tan θ < 0$
$\sin θ =$
(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)