(a+ib )(c+id) (e+if )(g+ih)=A+iB
Then prove that
tan^-1 b/a +tan ^-1 d/c +tan^-1f/e +tan ^-1h/g = tan^-1B/A

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \left(\mathrm{a}+\mathrm{ib}\right)\left(\mathrm{c}+\mathrm{id}\right)\left(\mathrm{e}+\mathrm{if}\right)\left(\mathrm{g}+\mathrm{ih}\right)=\mathrm{A}+\mathrm{iB} \\ \Rightarrow \arg \left\{\left(\mathrm{a}+\mathrm{ib}\right)\left(\mathrm{c}+\mathrm{id}\right)\left(\mathrm{e}+\mathrm{if}\right)\left(\mathrm{g}+\mathrm{ih}\right)\right\}=\arg \left(\mathrm{A}+\mathrm{iB}\right) ; \mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{Where} \arg \left(\mathrm{z}\right) \mathrm{denotes} \mathrm{argument} \mathrm{ofz}. \\ \mathrm{Now} \mathrm{we} \mathrm{know} \mathrm{that} \\ \mathbf{arg}\left({\mathbf{z}}_{\mathbf{1}}\mathbf{.}{\mathbf{z}}_{\mathbf{2}}\mathbf{.}{\mathbf{z}}_{\mathbf{3}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{z}}_{\mathbf{n}}\right)\mathbf{=}\mathbf{arg}\left({\mathbf{z}}_{\mathbf{1}}\right)\mathbf{+}\mathbf{arg}\left({\mathbf{z}}_{\mathbf{2}}\right)\mathbf{+}\mathbf{arg}\left({\mathbf{z}}_{\mathbf{3}}\right)\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{arg}\left({\mathbf{z}}_{\mathbf{n}}\right) \\ \mathrm{Hence} \mathrm{equation}\left(\mathrm{i}\right) \mathrm{becomes}, \\ \arg \left(\mathrm{a}+\mathrm{ib}\right)+\arg \left(\mathrm{c}+\mathrm{id}\right)+\arg \left(\mathrm{e}+\mathrm{if}\right)+\arg \left(\mathrm{g}+\mathrm{ih}\right)=\arg \left(\mathrm{A}+\mathrm{iB}\right) \\ \mathbf{\Rightarrow }{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{b}}{\mathbf{a}}\right)\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{d}}{\mathbf{c}}\right)\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{f}}{\mathbf{e}}\right)\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{h}}{\mathbf{g}}\right)\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{B}}{\mathbf{A}}\right)\mathbf{ }\left[\mathbf{Hence}\mathbf{ }\mathbf{Proved}\right] \end{gathered}$$

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