If sin A = frac{4}{5} and cos B = - frac{12}{ | vedclass

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(D) Given: $\sin A = \frac{4}{5}$ ($A$ is in the first quadrant,so $\cos A$ is positive) and $\cos B = -\frac{12}{13}$ ($B$ is in the third quadrant,s

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$-\frac{16}{65}$

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(D) Given: $\sin A = \frac{4}{5}$ ($A$ is in the first quadrant,so $\cos A$ is positive) and $\cos B = -\frac{12}{13}$ ($B$ is in the third quadrant,so $\sin B$ is negative).Step $1$: Find $\cos A$.$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.Step $2$: Find $\sin B$.$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (-\frac{12}{13})^2} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.Step $3$: Use the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$.$\cos(A + B) = (\frac{3}{5})(-\frac{12}{13}) - (\frac{4}{5})(-\frac{5}{13})$$= -\frac{36}{65} + \frac{20}{65}$$= -\frac{16}{65}$.

If $\sin A = \frac{4}{5}$ and $\cos B = - \frac{12}{13},$ where $A$ and $B$ lie in the first and third quadrants respectively,then $\cos (A + B) = $

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