If sin A = frac{4}{5} and cos B = -frac{12}{1 | vedclass

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(D) Given $\sin A = \frac{4}{5}$ and $\cos B = -\frac{12}{13}$.Since $A$ is in the first quadrant,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{16}{

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$-\frac{16}{65}$

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(D) Given $\sin A = \frac{4}{5}$ and $\cos B = -\frac{12}{13}$.Since $A$ is in the first quadrant,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$.Since $B$ is in the third quadrant,$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}$.Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:$\cos(A + B) = \left(\frac{3}{5}\right)\left(-\frac{12}{13}\right) - \left(\frac{4}{5}\right)\left(-\frac{5}{13}\right)$$= -\frac{36}{65} + \frac{20}{65}$$= -\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 quadrant respectively,then $\cos(A + B) = $

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