$\text{ If }\mathrm{\theta }={\tan }^{-1}\left(\mathrm{a}\right),\mathrm{\varphi }={\tan }^{-1}\left(\mathrm{b}\right)\text{ and }\mathrm{ab}=-1\text{ then }\mathrm{\theta }-\mathrm{\varphi }\text{ is equal to }$

$$\begin{gathered} Now \theta -\varphi ={\tan }^{-1}\left(a\right)-{\tan }^{-1}\left(b\right) \\ ={\tan }^{-1}\left(\frac{a-b}{1+ab}\right) \\ = {\tan }^{-1}(\frac{a-b}{0} ) \\ ={\tan }^{-1}(\infty ) \\ =\frac{\pi }{2} \end{gathered}$$