Question

If $\mathrm{A}$ lies in the third quadrant and $3\tan \left(\mathrm{A}\right)-4=0$ then $5\sin \left(2\mathrm{A}\right)+3\sin \left(\mathrm{A}\right)+4\cos \left(\mathrm{A}\right)$ is equal to?

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

Answer 1

The correct option is A

$0$

Step 1: Find $\cos \left(\mathrm{A}\right)$ and $\sin \left(\mathrm{A}\right)$

It is given that,

$$\begin{gathered} 3\tan \left(\mathrm{A}\right)-4=0 \\ \Rightarrow \tan \left(\mathrm{A}\right)=\frac{4}{3} \end{gathered}$$

Now,

$$\begin{gathered} {\tan }^2\left(\mathrm{A}\right)=\frac{16}{9} \\ {\sec }^2\left(\mathrm{A}\right)=1+\frac{16}{9}[{\sec }^2\left(\mathrm{A}\right)=1+{\tan }^2\left(\mathrm{A}\right)] \\ \Rightarrow {\sec }^2\left(\mathrm{A}\right)=\frac{25}{9} \end{gathered}$$

Since $\mathrm{A}$ lies in third quadrant.

$\sec \left(\mathrm{A}\right)=\frac{-5}{3}$and $\cos \left(\mathrm{A}\right)=\frac{-3}{5}$

$$\begin{gathered} {\sin }^2\left(\mathrm{A}\right)=1-\frac{9}{25}[{\sin }^2\left(\mathrm{A}\right)=1-{\cos }^2\left(\mathrm{A}\right)] \\ \Rightarrow {\sin }^2\left(\mathrm{A}\right)=\frac{16}{25} \\ \Rightarrow \sin \left(\mathrm{A}\right)=\frac{-4}{5}\left[\because \mathrm{In}{3}^{\mathrm{rd}}\mathrm{Quadrant}\sin \left(\mathrm{\theta }\right)\mathrm{is}\mathrm{negative}\right] \end{gathered}$$

Step 2: Find the value of the given expression

$$\begin{gathered} 5\left[2\sin \left(\mathrm{A}\right)\cos \left(\mathrm{A}\right)\right]+3\sin \left(\mathrm{A}\right)+4\cos \left(\mathrm{A}\right)[\sin \left(2\mathrm{A}\right)=2\sin \left(\mathrm{A}\right)\cos \left(\mathrm{A}\right)] \\ =5\left[2\times \left(\frac{-4}{5}\right)\times \left(\frac{-3}{5}\right)\right]+3\times \left(\frac{-4}{5}\right)\mathbf{+}4\times \left(\frac{-3}{5}\right) \\ =0 \end{gathered}$$

Thus, $\mathrm{Option}\left(\mathrm{A}\right)$ is correct.