Question

The value of $\tan (\alpha+\beta)$, given that $\cot \,\alpha=\frac{1}{2}$, $\alpha \in\left(\pi, \frac{3 \pi}{2}\right)$ and $\sec \,\beta=\frac{-5}{3}, \beta \in\left(\frac{\pi}{2}, \pi\right)$ is

• A. $\frac{1}{11}$
• B. $\frac{2}{11}$
• C. $\frac{5}{11}$
• D. $\frac{3}{11}$

Answer 1

Answer: (B)

Given, $\cot \alpha=\frac{1}{2} \Rightarrow \tan \alpha=2$ and $\sec \beta=\frac{-5}{3}$
Then, $\tan \beta=\sqrt{\sec ^{2} \beta-1} \Rightarrow \tan \beta=\pm \sqrt{\frac{25}{9}-1}=\pm \sqrt{\frac{16}{9}}$
$\Rightarrow \tan \beta=\pm \frac{4}{3}$ But, $\tan \beta=\frac{-4}{3}$
$\left[\because \tan \beta\right.$ is negative in $II ^{ Id }$ quadrant $]$
$\therefore \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\frac{2+\left(-\frac{4}{3}\right)}{1-(2)\left(\frac{-4}{3}\right)}=\frac{2}{11} $.