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, $\tan \beta =\frac{-4}{3}$
$[\because \tan \beta$ is negative in $I{I}^{Id}$ quadrant $]$
$\therefore \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \cdot \tan \beta }=\frac{2+\left(-\frac{4}{3}\right)}{1-(2)\left(\frac{-4}{3}\right)}=\frac{2}{11}$.