Question

If $\cot \alpha =\frac{1}{2},sec\beta =-\frac{5}{3}$, where $\pi <\alpha <\frac{3\pi }{2}$ and $\frac{\pi }{2}<\beta <\pi$. Find the value of $\tan (\alpha +\beta )$. State the quadrant in which $\alpha +\beta$ terminate.

Answer 1

$\cot \alpha =\frac{1}{2}$

$\Rightarrow \tan \alpha =2$

$1+{\tan }^2\beta ={\sec }^2\beta$

$\Rightarrow 1+{\tan }^2\beta =\frac{25}{9}$

$\Rightarrow {\tan }^2\beta =\frac{25}{9}-1$

$\Rightarrow {\tan }^2\beta =\frac{16}{9}$

$\Rightarrow \tan \beta =-\frac{4}{3}\left(\frac{\pi }{2}<\beta <\pi \right)$

$\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

$=\frac{2-\frac{4}{3}}{1-2\times -\frac{4}{3}}=\frac{6-4}{3+8}$

$=\frac{2}{11}$

$\pi +\frac{\pi }{2}<\alpha +\beta <\frac{3\pi }{2}+\pi$

$\Rightarrow \frac{3\pi }{2}<\alpha +\beta <\frac{5\pi }{2}$

$\Rightarrow \alpha +\beta \in \left(\frac{3\pi }{2},\frac{5\pi }{2}\right)$

since $\tan \left(\alpha +\beta \right)$ is positive . So , $\alpha +\beta \in \left(2\pi ,\frac{5\pi }{2}\right)$ i.e . first quadrant