If sin A = frac{-7}{25},cos B = frac{8}{17},A | vedclass

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(B) Given $\sin A = \frac{-7}{25}$. Since $\sin A < 0$ and $A$ is not in the $3^{	ext{rd}}$ quadrant,$A$ must be in the $4^{	ext{th}}$ quadrant.In t

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$\frac{1}{3}$

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(B) Given $\sin A = \frac{-7}{25}$. Since $\sin A In the $4^{	ext{th}}$ quadrant,$\cos A > 0$. Thus,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - (\frac{-7}{25})^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}$.Therefore,$	an A = \frac{\sin A}{\cos A} = \frac{-7/25}{24/25} = \frac{-7}{24}$.Given $\cos B = \frac{8}{17}$. Since $\cos B > 0$ and $B$ is not in the $1^{	ext{st}}$ quadrant,$B$ must be in the $4^{	ext{th}}$ quadrant.In the $4^{	ext{th}}$ quadrant,$\sin B Therefore,$\cot B = \frac{\cos B}{\sin B} = \frac{8/17}{-15/17} = \frac{-8}{15}$.Now,$8 	an A - 5 \cot B = 8(\frac{-7}{24}) - 5(\frac{-8}{15}) = \frac{-7}{3} + \frac{8}{3} = \frac{1}{3}$.

If $\sin A = \frac{-7}{25}$,$\cos B = \frac{8}{17}$,$A$ does not lie in the $3^{\text{rd}}$ quadrant and $B$ does not lie in the $1^{\text{st}}$ quadrant,then $8 \tan A - 5 \cot B =$

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