Question

If $\tan \left(\theta \right)=-\frac{4}{3}$, then $\sin \left(\theta \right)$ is

• A. $-\frac{4}{5}$ but not $\frac{4}{5}$
• B. $-\frac{4}{5}$ or $\frac{4}{5}$
• C. $\frac{4}{5}$ but not $-\frac{4}{5}$
• D. none of these

Answer 1

The correct option is B

$-\frac{4}{5}$ or $\frac{4}{5}$

Explanation for the correct option

Given: $\tan \left(\theta \right)=-\frac{4}{3}$

$$\begin{gathered} \Rightarrow \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}=-\frac{4}{3}\left[\because \tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\right] \\ \Rightarrow \frac{\sin \left(\theta \right)}{\sqrt{1-{\sin }^2\left(\theta \right)}}=-\frac{4}{3}\left[\because {\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1\right] \end{gathered}$$

squaring both sides

$$\begin{gathered} \Rightarrow \frac{{\sin }^2\left(\theta \right)}{{\left(\sqrt{1-{\sin }^2\left(\theta \right)}\right)}^2}={\left(-\frac{4}{3}\right)}^2 \\ \Rightarrow \frac{{\sin }^2\left(\theta \right)}{1-{\sin }^2\left(\theta \right)}=\frac{16}{9} \\ \Rightarrow 16-16{\sin }^2\left(\theta \right)=9{\sin }^2\left(\theta \right) \\ \Rightarrow 16{\sin }^2\left(\theta \right)+9{\sin }^2\left(\theta \right)=16 \\ \Rightarrow 25{\sin }^2\left(\theta \right)=16 \\ \Rightarrow {\sin }^2\left(\theta \right)=\frac{16}{25} \\ \Rightarrow \sin \left(\theta \right)=\pm \frac{4}{5} \end{gathered}$$

Hence option B is correct.