Answer
$\tan^{-1} (\dfrac{33}{x})- \tan^{-1} (\dfrac{8}{x})$
Work Step by Step
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
$\theta=\alpha+\theta-\alpha= \tan^{-1} (\dfrac{33}{x})- \tan^{-1} (\dfrac{8}{x})$
