Question Number 124501 by mnjuly1970 last updated on 03/Dec/20
$$\:\:\:\:\:\:\:\:\:\:….{nice}\:\:{calculus}.. \\ $$$$\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\::\:{sin}^{\mathrm{2}} \left({A}\right)+{sin}^{\mathrm{2}} \left({B}\right)+{sin}^{\mathrm{2}} \left({C}\right)=\mathrm{2} \\ $$$${prove}\:{that}:\:{A}\overset{\Delta} {{B}C}\:{is}\:{right}\:{triangle} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathscr{G}{ood}\:{luck}. \\ $$
Answered by mindispower last updated on 03/Dec/20
$${sin}\left({c}\right)={sin}\left({a}+{b}\right) \\ $$$$\left.\Leftrightarrow{sin}^{\mathrm{2}} \left({A}\right)+{sin}^{\mathrm{2}} \left({B}\right)+\left({sin}\left({A}+{B}\right)\right)^{\mathrm{2}} \right)=\mathrm{2} \\ $$$${sin}^{\mathrm{2}} \left({A}+{B}\right)={sin}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)+{cos}^{\mathrm{2}} \left({A}\right){sin}^{\mathrm{2}} \left({B}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{A}\right){sin}\left(\mathrm{2}{B}\right) \\ $$$${sin}^{\mathrm{2}} \left({A}\right)−\mathrm{1}+{sin}^{\mathrm{2}} \left({B}\right)−\mathrm{1}+\frac{{sin}\left(\mathrm{2}{A}\right){sin}\left(\mathrm{2}{B}\right)}{\mathrm{2}}+ \\ $$$${sin}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)+{sin}^{\mathrm{2}} \left({B}\right){cos}^{\mathrm{2}} \left({A}\right)=\mathrm{0} \\ $$$$\Leftrightarrow−{cos}^{\mathrm{2}} \left({A}\right)+{cos}^{\mathrm{2}} \left({A}\right){sin}^{\mathrm{2}} \left({B}\right)−{cos}^{\mathrm{2}} \left({B}\right)+{cos}^{\mathrm{2}} \left({B}\right){sin}^{\mathrm{2}} \left({A}\right) \\ $$$$+\frac{{sin}\left(\mathrm{2}{A}\right){sin}\left(\mathrm{2}{B}\right)}{\mathrm{2}}=\mathrm{0} \\ $$$$\Leftrightarrow−\mathrm{2}{cos}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)+\mathrm{2}{sin}\left({A}\right){sin}\left({B}\right){cos}\left({A}\right){cos}\left({B}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}{cos}\left({A}\right){cos}\left({B}\right)\left({sin}\left({A}\right){sin}\left({B}\right)−{cos}\left({A}\right){cos}\left({B}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{cos}\left({A}\right){cos}\left({B}\right){cos}\left({A}+{B}\right)=\mathrm{0} \\ $$$$\Rightarrow{A}=\frac{\pi}{\mathrm{2}},{B}=\frac{\pi}{\mathrm{2}},{or}\:{A}+{B}=\frac{\pi}{\mathrm{2}}\Rightarrow{C}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 03/Dec/20
$${thank}\:{you}\:{so}\:{much} \\ $$$${mr}\:{mindspower} \\ $$$${nice}\:{as}\:{always}… \\ $$
Answered by Dwaipayan Shikari last updated on 03/Dec/20
$$\mathrm{2}{sin}^{\mathrm{2}} {A}+\mathrm{2}{sin}^{\mathrm{2}} {B}+\mathrm{2}{sin}^{\mathrm{2}} {C}=\mathrm{4} \\ $$$$\mathrm{1}−{cos}\mathrm{2}{A}+\mathrm{1}−{cos}\mathrm{2}{B}+\mathrm{1}−{sin}\mathrm{2}{C}=\mathrm{4} \\ $$$${cos}\mathrm{2}{A}+{cos}\mathrm{2}{B}+{cos}\mathrm{2}{C}=−\mathrm{1} \\ $$$$\mathrm{2}{cos}\left({A}+{B}\right){cos}\left({A}−{B}\right)+{cos}\mathrm{2}{C}=−\mathrm{1} \\ $$$$\mathrm{2}{cos}\left({A}+{B}\right){cos}\left({A}−{B}\right)+\mathrm{2}{cos}^{\mathrm{2}} {C}=\mathrm{0} \\ $$$${cos}\left({A}+{B}\right){cos}\left({A}−{B}\right)+{cos}^{\mathrm{2}} {C}=\mathrm{0} \\ $$$${cosC}\:{cos}\left({A}−{B}\right)={cos}^{\mathrm{2}} {C} \\ $$$${cosC}\left({cosC}−{cos}\left({A}−{B}\right)\right)=\mathrm{0} \\ $$$${cosC}=\mathrm{0}\:\:{C}=\frac{\pi}{\mathrm{2}}\:\:\:\:\:\:{or}\:\:{cosC}={cos}\left({A}−{B}\right)\:\Rightarrow{C}={A}−{B}\Rightarrow{C}+{B}={A} \\ $$
Commented by mnjuly1970 last updated on 03/Dec/20
$${thank}\:{you}\:{so}\:{much} \\ $$$${extraordinary}… \\ $$$$ \\ $$