If A lies in the second quadrant and 3tan A + | vedclass

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(D) Given $3	an A + 4 = 0,$ so $	an A = -\frac{4}{3}$.Since $A$ lies in the second quadrant,$	an A$ is negative,$\sin A$ is positive,and $\cos A$ i

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$\frac{23}{10}$

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(D) Given $3	an A + 4 = 0,$ so $	an A = -\frac{4}{3}$.Since $A$ lies in the second quadrant,$	an A$ is negative,$\sin A$ is positive,and $\cos A$ is negative.Using the identity $\sec^2 A = 1 + 	an^2 A = 1 + (-\frac{4}{3})^2 = 1 + \frac{16}{9} = \frac{25}{9}$,we get $\sec A = -\frac{5}{3}$ (as $A$ is in the second quadrant).Thus,$\cos A = \frac{1}{\sec A} = -\frac{3}{5}$.Also,$\sin A = 	an A 	imes \cos A = (-\frac{4}{3}) 	imes (-\frac{3}{5}) = \frac{4}{5}$.And $\cot A = \frac{1}{	an A} = -\frac{3}{4}$.Substituting these values into the expression $2\cot A - 5\cos A + \sin A$:$= 2(-\frac{3}{4}) - 5(-\frac{3}{5}) + \frac{4}{5}$$= -\frac{3}{2} + 3 + \frac{4}{5}$$= \frac{-15 + 30 + 8}{10} = \frac{23}{10}$.

If $A$ lies in the second quadrant and $3\tan A + 4 = 0,$ the value of $2\cot A - 5\cos A + \sin A$ is equal to

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