Question

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

• A. 0
• B. $\frac{-24}{5}$
• C. $\frac{24}{5}$
• D. $\frac{48}{5}$

Answer 1

The correct option is A 0

$3\tan A=4\Rightarrow \tan A=\frac{4}{3}$

${\tan }^{2}A=\frac{16}{9},{\sec }^{2}A={\tan }^2+1=\frac{16}{9}+1=\frac{25}{9}$

$\sec A=\frac{-5}{3},\cos A=\frac{-3}{5}$ [since A lies in the third quadrant]

${\sin }^{2}A=1-{\cos }^{2}A=1-\frac{9}{25}=\frac{16}{25}\Rightarrow \sin A=\frac{-4}{5}$

so $\sin A=-\frac{4}{5}$ and $\cos A=-\frac{3}{5}$

$\Rightarrow 5\times 2\sin A\cos A+3\sin A+4\cos A$

$\Rightarrow 5\times 2-\frac{4}{5}\times \frac{3}{5}+3\times \frac{4}{5}+4\times \frac{3}{5}$

$\Rightarrow 10\times \frac{12}{15}-\frac{12}{5}-\frac{12}{5}=0$