Without using tables, find $\;$ $\tan \left(\alpha - \dfrac{\pi}{4}\right)$, if $\cos \alpha = \dfrac{-9}{41}$, $\pi < \alpha < \dfrac{3 \pi}{2}$
Given: $\;$ $\pi < \alpha < \dfrac{3 \pi}{2}$
$\implies$ $\alpha$ lies in the third quadrant.
$\implies$ Both $\sin \alpha$, $\cos \alpha$ are negative and $\tan \alpha$ is positive.
Given: $\;$ $\cos \alpha = \dfrac{-9}{41}$
$\therefore \;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(\dfrac{-9}{41}\right)^2} = \dfrac{-40}{41}$
$\therefore \;$ $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha} = \dfrac{-40 / 41}{-9 / 41} = \dfrac{40}{9}$
Now, $\;$ $\tan \left(\alpha - \dfrac{\pi}{4}\right) = \dfrac{\tan \alpha - \tan \dfrac{\pi}{4}}{1 - \tan \alpha \tan \dfrac{\pi}{4}} = \dfrac{\dfrac{40}{9} - 1}{1 + \dfrac{40}{9}} = \dfrac{31}{49}$