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Question Number 5313 by qw last updated on 06/May/16

tan^(−1) (√((a(a+b+c))/(bc))) + tan^(−1) (√((b(a+b+c))/(ca))) + tan^(−1) (√((c(a+b+c))/(ab))) =

$$\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{a}\left({a}+{b}+{c}\right)}{{bc}}}\:+\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{b}\left({a}+{b}+{c}\right)}{{ca}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{c}\left({a}+{b}+{c}\right)}{{ab}}}\:\:= \\ $$

Commented by Yozzii last updated on 07/May/16

Let j=tan(tan^(−1) u+tan^(−1) v+tan^(−1) w) j=((tan(tan^(−1) u)+tan(tan^(−1) v+tan^(−1) w))/(1−{tan(tan^(−1) u)}{tan(tan^(−1) v+tan^(−1) w)})) Assuming principal answers only for tan(tan^(−1) x)=x, j=((u+((tan(tan^(−1) v)+tan(tan^(−1) w))/(1−{tan(tan^(−1) v)}{tan(tan^(−1) w)})))/(1−u×((tan(tan^(−1) v)+tan(tan^(−1) w))/(1−{tan(tan^(−1) v)}{tan(tan^(−1) w)})))) j=((u+((v+w)/(1−vw)))/(1−u((v+w)/(1−vw)))) j=((u+v+w−uvw)/(1−vw−uv−uw)) ⇒tan^(−1) j=tan^(−1) u+tan^(−1) v+tan^(−1) w=tan^(−1) ((u+v+w−uvw)/(1−(vw+uv+uw))) or more generally, Σ_(i=1) ^3 tan^(−1) a_i =nπ+tan^(−1) ((a_1 +a_2 +a_3 −a_1 a_2 a_3 )/(1−(a_1 a_2 +a_2 a_3 +a_3 a_1 ))) (n∈Z). Let s=(√(a+b+c)) , u=(√((a(a+b+c))/(bc))), v=(√((b(a+b+c))/(ca))), w=(√((c(a+b+c))/(ab))) with nonzero a,b,c. ∴u+v+w=s((√(a/(bc)))+(√(b/(ca)))+(√(c/(ab))))=((s(∣a∣+∣b∣+∣c∣)(√(abc)))/(∣abc∣)) and uvw=(√(((a(a+b+c))/(bc))×((b(a+b+c))/(ca))×((c(a+b+c))/(ab)))) uwv=(√((abc(a+b+c)(a+b+c)(a+b+c))/(a^2 b^2 c^2 )))=(s^3 /(√(abc))) uv+vw+wu=((s^2 (∣ab∣+∣bc∣+∣ca∣))/(∣abc∣)) ∴tan^(−1) j=nπ+tan^(−1) ((((s(∣a∣+∣b∣+∣c∣)(√(abc)))/(∣abc∣))−(s^3 /(√(abc))))/(1−((s^2 (∣ab∣+∣bc∣+∣ca∣))/(∣abc∣)))) tan^(−1) j=nπ+tan^(−1) ((s(∣a∣+∣b∣+∣c∣)(√(abc))−((s^3 ∣abc∣)/(√(abc))))/(∣abc∣−s^2 (∣ab∣+∣bc∣+∣ca∣))) tan^(−1) u+tan^(−1) v+tan^(−1) w=(((√(abc(a+b+v)))(∣a∣+∣b∣+∣c∣)−(a+b+c)^(3/2) (√((a^2 b^2 c^2 )/(abc))))/(∣abc∣−(a+b+c)(∣ab∣+∣bc∣+∣ca∣)))+nπ tan^(−1) u+tan^(−1) v+tan^(−1) w=(((√(abc(a+b+c)))(∣a∣+∣b∣+∣c∣)−(√(abc(a+b+c)))(a+b+c))/(∣abc∣−(a+b+c)(∣ab∣+∣bc∣+∣ca∣)))+nπ tan^(−1) u+tan^(−1) v+tan^(−1) w=(((√(abc(a+b+c){))∣a∣−a+∣b∣−b+∣c∣−c})/(∣abc∣−(a+b+c)(∣ab∣+∣bc∣+∣ca∣)))+nπ If a,b,c≥0⇒∣a∣=a,∣b∣=b,∣c∣=c⇒tan^(−1) u+tan^(−1) v+tan^(−1) w=nπ.

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

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