## Precalculus (6th Edition) Blitzer

$\tan^{-1} (\dfrac{33}{x})- \tan^{-1} (\dfrac{8}{x})$
Using problem 93, we get: Suppose $\tan (\alpha+\theta)=\dfrac{33}{x}$ This gives: $\alpha+\theta= \tan^{-1} \dfrac{33}{x}$ And also: $\tan \alpha=\dfrac{8}{x}$ Which gives: $\alpha= \tan^{-1} \dfrac{8}{x}$ Therefore, we have $\theta=\alpha+\theta-\alpha= \tan^{-1} (\dfrac{33}{x})- \tan^{-1} (\dfrac{8}{x})$