Question Number 189941 by Rupesh123 last updated on 24/Mar/23
Answered by Frix last updated on 24/Mar/23
$$\mathrm{2arctan}\:{x}\:+\mathrm{arctan}\:{y}\:=\mathrm{arctan}\:\frac{{x}^{\mathrm{2}} {y}−\mathrm{2}{x}−{y}}{{x}^{\mathrm{2}} +\mathrm{2}{xy}−\mathrm{1}} \\ $$$$\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$
Commented by Rupesh123 last updated on 24/Mar/23
Nice solution, sir!
Answered by CElcedricjunior last updated on 26/Mar/23
![2.arctan(((b+c)/a))+arctan(((2a(b+c))/((b+c)^2 −a^2 )))=? tan(2𝛉+𝛃)=((tan2𝛉+tan𝛃)/(1−tan2𝛉tan𝛃)) =((((2tan𝛉)/(1−tan^2 𝛉))+tan𝛃)/(1−((2tan𝛉tan𝛃)/(1−tan^2 𝛉)))) =((2tan𝛉+tan𝛃−tan^2 𝛉×tan𝛃)/(1−tan^2 𝛉−2tan𝛉×tan𝛃)) =((2(((b+c)/a))+((2a(b+c))/((b+c)^2 −a^2 ))−(((b+c)/a))^2 (((2a(b+c))/((b+c)^2 −a^2 ))))/(1−(((b+c)^2 )/a^2 )−2(((b+c))/a)(((2a(b+c)))/([(b+c)^2 −a^2 ])))) =(((([2a(b+c)][(b+c)^2 −a^2 ]+2a^3 (b+c)−2a(b+c)^2 )/(a[(b+c)^2 −a^2 ])) )/((a^2 (b+c)^2 −a^4 −(b+c)^4 +a^2 (b+c)^2 −4a^2 (b+c)^2 )/(a^2 [(b+c)^2 −a^2 ]))) =((2a^2 (b+c)^2 [(b+c)−1))/(−a^4 −(b+c)^4 −2a^2 (b+c)^2 )) =((2a^2 (b+c)^2 [(b+c)^2 −1])/((a^2 +(b+c)^2 )^2 )) =](https://www.tinkutara.com/question/Q190077.png)
$$\mathrm{2}.\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{b}}+\boldsymbol{{c}}}{\boldsymbol{{a}}}\right)+\boldsymbol{{arctan}}\left(\frac{\mathrm{2}\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)}{\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\right)=? \\ $$$$\boldsymbol{{tan}}\left(\mathrm{2}\boldsymbol{\theta}+\boldsymbol{\beta}\right)=\frac{\boldsymbol{{tan}}\mathrm{2}\boldsymbol{\theta}+\boldsymbol{{tan}\beta}}{\mathrm{1}−\boldsymbol{{tan}}\mathrm{2}\boldsymbol{\theta{tan}\beta}} \\ $$$$=\frac{\frac{\mathrm{2}\boldsymbol{{tan}\theta}}{\mathrm{1}−\boldsymbol{{ta}}\overset{\mathrm{2}} {\boldsymbol{{n}}\theta}}+\boldsymbol{{tan}\beta}}{\mathrm{1}−\frac{\mathrm{2}\boldsymbol{{tan}\theta{tan}\beta}}{\mathrm{1}−\boldsymbol{{ta}}\overset{\mathrm{2}} {\boldsymbol{{n}}\theta}}} \\ $$$$=\frac{\mathrm{2}\boldsymbol{{tan}\theta}+\boldsymbol{{tan}\beta}−\boldsymbol{{ta}}\overset{\mathrm{2}} {\boldsymbol{{n}}\theta}×\boldsymbol{{tan}\beta}}{\mathrm{1}−\boldsymbol{{ta}}\overset{\mathrm{2}} {\boldsymbol{{n}}\theta}−\mathrm{2}\boldsymbol{{tan}\theta}×\boldsymbol{{tan}\beta}} \\ $$$$=\frac{\mathrm{2}\left(\frac{\boldsymbol{{b}}+\boldsymbol{{c}}}{\boldsymbol{{a}}}\right)+\frac{\mathrm{2}\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)}{\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }−\left(\frac{\boldsymbol{{b}}+\boldsymbol{{c}}}{\boldsymbol{{a}}}\right)^{\mathrm{2}} \left(\frac{\mathrm{2}\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)}{\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\right)}{\mathrm{1}−\frac{\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} }−\mathrm{2}\frac{\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)}{\boldsymbol{{a}}}\frac{\left(\mathrm{2}\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)\right)}{\left[\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right]}} \\ $$$$=\frac{\frac{\left[\mathrm{2}\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)\right]\left[\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right]+\mathrm{2}\boldsymbol{{a}}^{\mathrm{3}} \left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)−\mathrm{2}\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{a}}\left[\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right]}\:\:}{\frac{\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{4}} −\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{4}} +\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\mathrm{4}\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} \left[\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right]}} \\ $$$$=\frac{\mathrm{2}\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} \left[\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)−\mathrm{1}\right)}{−\boldsymbol{{a}}^{\mathrm{4}} −\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{4}} −\mathrm{2}\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} \left[\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} −\mathrm{1}\right]}{\left(\boldsymbol{{a}}^{\mathrm{2}} +\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} \right)^{\mathrm{2}} }\: \\ $$$$= \\ $$