Question Number 5313 by qw last updated on 06/May/16
$$\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{a}\left({a}+{b}+{c}\right)}{{bc}}}\:+\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{b}\left({a}+{b}+{c}\right)}{{ca}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{c}\left({a}+{b}+{c}\right)}{{ab}}}\:\:= \\ $$
Commented by Yozzii last updated on 07/May/16
$${Let}\:{j}={tan}\left({tan}^{−\mathrm{1}} {u}+{tan}^{−\mathrm{1}} {v}+{tan}^{−\mathrm{1}} {w}\right) \\ $$$${j}=\frac{{tan}\left({tan}^{−\mathrm{1}} {u}\right)+{tan}\left({tan}^{−\mathrm{1}} {v}+{tan}^{−\mathrm{1}} {w}\right)}{\mathrm{1}−\left\{{tan}\left({tan}^{−\mathrm{1}} {u}\right)\right\}\left\{{tan}\left({tan}^{−\mathrm{1}} {v}+{tan}^{−\mathrm{1}} {w}\right)\right\}} \\ $$$${Assuming}\:{principal}\:{answers}\:{only}\:{for}\:{tan}\left({tan}^{−\mathrm{1}} {x}\right)={x}, \\ $$$${j}=\frac{{u}+\frac{{tan}\left({tan}^{−\mathrm{1}} {v}\right)+{tan}\left({tan}^{−\mathrm{1}} {w}\right)}{\mathrm{1}−\left\{{tan}\left({tan}^{−\mathrm{1}} {v}\right)\right\}\left\{{tan}\left({tan}^{−\mathrm{1}} {w}\right)\right\}}}{\mathrm{1}−{u}×\frac{{tan}\left({tan}^{−\mathrm{1}} {v}\right)+{tan}\left({tan}^{−\mathrm{1}} {w}\right)}{\mathrm{1}−\left\{{tan}\left({tan}^{−\mathrm{1}} {v}\right)\right\}\left\{{tan}\left({tan}^{−\mathrm{1}} {w}\right)\right\}}} \\ $$$${j}=\frac{{u}+\frac{{v}+{w}}{\mathrm{1}−{vw}}}{\mathrm{1}−{u}\frac{{v}+{w}}{\mathrm{1}−{vw}}} \\ $$$${j}=\frac{{u}+{v}+{w}−{uvw}}{\mathrm{1}−{vw}−{uv}−{uw}} \\ $$$$\Rightarrow{tan}^{−\mathrm{1}} {j}={tan}^{−\mathrm{1}} {u}+{tan}^{−\mathrm{1}} {v}+{tan}^{−\mathrm{1}} {w}={tan}^{−\mathrm{1}} \frac{{u}+{v}+{w}−{uvw}}{\mathrm{1}−\left({vw}+{uv}+{uw}\right)} \\ $$$${or}\:{more}\:{generally},\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{tan}^{−\mathrm{1}} {a}_{{i}} ={n}\pi+{tan}^{−\mathrm{1}} \frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} −{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} }{\mathrm{1}−\left({a}_{\mathrm{1}} {a}_{\mathrm{2}} +{a}_{\mathrm{2}} {a}_{\mathrm{3}} +{a}_{\mathrm{3}} {a}_{\mathrm{1}} \right)}\:\:\left({n}\in\mathbb{Z}\right). \\ $$$${Let}\:{s}=\sqrt{{a}+{b}+{c}}\:,\:{u}=\sqrt{\frac{{a}\left({a}+{b}+{c}\right)}{{bc}}}, \\ $$$${v}=\sqrt{\frac{{b}\left({a}+{b}+{c}\right)}{{ca}}},\:{w}=\sqrt{\frac{{c}\left({a}+{b}+{c}\right)}{{ab}}} \\ $$$${with}\:{nonzero}\:{a},{b},{c}. \\ $$$$\therefore{u}+{v}+{w}={s}\left(\sqrt{\frac{{a}}{{bc}}}+\sqrt{\frac{{b}}{{ca}}}+\sqrt{\frac{{c}}{{ab}}}\right)=\frac{{s}\left(\mid{a}\mid+\mid{b}\mid+\mid{c}\mid\right)\sqrt{{abc}}}{\mid{abc}\mid}\: \\ $$$${and}\:{uvw}=\sqrt{\frac{{a}\left({a}+{b}+{c}\right)}{{bc}}×\frac{{b}\left({a}+{b}+{c}\right)}{{ca}}×\frac{{c}\left({a}+{b}+{c}\right)}{{ab}}} \\ $$$${uwv}=\sqrt{\frac{{abc}\left({a}+{b}+{c}\right)\left({a}+{b}+{c}\right)\left({a}+{b}+{c}\right)}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }}=\frac{{s}^{\mathrm{3}} }{\:\sqrt{{abc}}} \\ $$$${uv}+{vw}+{wu}=\frac{{s}^{\mathrm{2}} \left(\mid{ab}\mid+\mid{bc}\mid+\mid{ca}\mid\right)}{\mid{abc}\mid} \\ $$$$\therefore{tan}^{−\mathrm{1}} {j}={n}\pi+{tan}^{−\mathrm{1}} \frac{\frac{{s}\left(\mid{a}\mid+\mid{b}\mid+\mid{c}\mid\right)\sqrt{{abc}}}{\mid{abc}\mid}−\frac{{s}^{\mathrm{3}} }{\:\sqrt{{abc}}}}{\mathrm{1}−\frac{{s}^{\mathrm{2}} \left(\mid{ab}\mid+\mid{bc}\mid+\mid{ca}\mid\right)}{\mid{abc}\mid}} \\ $$$${tan}^{−\mathrm{1}} {j}={n}\pi+{tan}^{−\mathrm{1}} \frac{{s}\left(\mid{a}\mid+\mid{b}\mid+\mid{c}\mid\right)\sqrt{{abc}}−\frac{{s}^{\mathrm{3}} \mid{abc}\mid}{\:\sqrt{{abc}}}}{\mid{abc}\mid−{s}^{\mathrm{2}} \left(\mid{ab}\mid+\mid{bc}\mid+\mid{ca}\mid\right)} \\ $$$${tan}^{−\mathrm{1}} {u}+{tan}^{−\mathrm{1}} {v}+{tan}^{−\mathrm{1}} {w}=\frac{\sqrt{{abc}\left({a}+{b}+{v}\right)}\left(\mid{a}\mid+\mid{b}\mid+\mid{c}\mid\right)−\left({a}+{b}+{c}\right)^{\mathrm{3}/\mathrm{2}} \sqrt{\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{abc}}}}{\mid{abc}\mid−\left({a}+{b}+{c}\right)\left(\mid{ab}\mid+\mid{bc}\mid+\mid{ca}\mid\right)}+{n}\pi \\ $$$${tan}^{−\mathrm{1}} {u}+{tan}^{−\mathrm{1}} {v}+{tan}^{−\mathrm{1}} {w}=\frac{\sqrt{{abc}\left({a}+{b}+{c}\right)}\left(\mid{a}\mid+\mid{b}\mid+\mid{c}\mid\right)−\sqrt{{abc}\left({a}+{b}+{c}\right)}\left({a}+{b}+{c}\right)}{\mid{abc}\mid−\left({a}+{b}+{c}\right)\left(\mid{ab}\mid+\mid{bc}\mid+\mid{ca}\mid\right)}+{n}\pi \\ $$$${tan}^{−\mathrm{1}} {u}+{tan}^{−\mathrm{1}} {v}+{tan}^{−\mathrm{1}} {w}=\frac{\left.\sqrt{{abc}\left({a}+{b}+{c}\right)\left\{\right.}\mid{a}\mid−{a}+\mid{b}\mid−{b}+\mid{c}\mid−{c}\right\}}{\mid{abc}\mid−\left({a}+{b}+{c}\right)\left(\mid{ab}\mid+\mid{bc}\mid+\mid{ca}\mid\right)}+{n}\pi \\ $$$$ \\ $$$${If}\:{a},{b},{c}\geqslant\mathrm{0}\Rightarrow\mid{a}\mid={a},\mid{b}\mid={b},\mid{c}\mid={c}\Rightarrow{tan}^{−\mathrm{1}} {u}+{tan}^{−\mathrm{1}} {v}+{tan}^{−\mathrm{1}} {w}={n}\pi. \\ $$