If A lies in the third quadrant and 3 tan A - | vedclass

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(A) Given $3 	an A - 4 = 0 \Rightarrow 	an A = \frac{4}{3}$.Since $A$ lies in the third quadrant,both $\sin A$ and $\cos A$ are negative.Using $	an

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(A) Given $3 	an A - 4 = 0 \Rightarrow 	an A = \frac{4}{3}$.Since $A$ lies in the third quadrant,both $\sin A$ and $\cos A$ are negative.Using $	an A = \frac{4}{3}$,we have $\sin A = -\frac{4}{5}$ and $\cos A = -\frac{3}{5}$.Now,$5 \sin 2A + 3 \sin A + 4 \cos A = 5(2 \sin A \cos A) + 3 \sin A + 4 \cos A$.Substituting the values: $10 \left(-\frac{4}{5}ight) \left(-\frac{3}{5}ight) + 3 \left(-\frac{4}{5}ight) + 4 \left(-\frac{3}{5}ight)$.$= 10 \left(\frac{12}{25}ight) - \frac{12}{5} - \frac{12}{5}$.$= \frac{24}{5} - \frac{24}{5} = 0$.

If $A$ lies in the third quadrant and $3 \tan A - 4 = 0,$ then $5 \sin 2A + 3 \sin A + 4 \cos A = $

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