Question Number 81884 by rajesh4661kumar@gmail.com last updated on 16/Feb/20
Commented by mind is power last updated on 16/Feb/20
$${applie} \\ $$$${we}\:{can}\:{find}\:{triangle}\: \\ $$$${withe}\:\:\:{sin}^{−} \:{x},{sin}^{−} {y},{sin}^{−} {z}\:{angle}\:\:{withe}\:{oriantation} \\ $$$${a}+{b}+{c}=\pi \\ $$$$\Rightarrow{sin}\left({a}\right){cos}\left({a}\right)+{sin}\left({b}\right){cos}\left({b}\right)+{sin}\left({c}\right){cos}\left({c}\right)=\mathrm{2}{sin}\left({a}\right){sin}\left({b}\right){sin}\left({c}\right)..\mathrm{1}? \\ $$$$\Rightarrow{sin}\left(\mathrm{2}{a}\right)+{sin}\left(\mathrm{2}{b}\right)+{sin}\left(\mathrm{2}{c}\right)=\mathrm{4}{sin}\left({a}\right){sin}\left({b}\right){sin}\left({c}\right)..\mathrm{2} \\ $$$${sin}\left(\mathrm{2}{a}\right)+{sin}\left(\mathrm{2}{b}\right)=\mathrm{2}{cos}\left({a}−{b}\right){sin}\left({a}+{b}\right) \\ $$$${sin}\left(\mathrm{2}{c}\right)={sin}\left(\mathrm{2}\pi−\mathrm{2}{a}−\mathrm{2}{b}\right)=−{sin}\left(\mathrm{2}{a}+\mathrm{2}{b}\right)=−\mathrm{2}{sin}\left({a}+{b}\right){cos}\left({a}+{b}\right) \\ $$$${sin}\left(\mathrm{2}{a}\right)+{sin}\left(\mathrm{2}{b}\right)+{sin}\left(\mathrm{2}{c}\right) \\ $$$$=\mathrm{2}{cos}\left({a}−{b}\right){sin}\left({a}+{b}\right)−\mathrm{2}{sin}\left({a}+{b}\right){cos}\left({a}+{b}\right) \\ $$$$=\mathrm{2}{sin}\left({a}+{b}\right)\left({cos}\left({a}−{b}\right)−{cos}\left({a}+{b}\right)\right) \\ $$$$=\mathrm{4}{sin}\left({a}\right){sin}\left({b}\right){sin}\left(\pi−{c}\right)=\mathrm{4}{sin}\left({a}\right){sin}\left({b}\right){sin}\left({c}\right) \\ $$$$\mathrm{2}..{proved}\:\:\Leftrightarrow\mathrm{1}\:{True} \\ $$$${for}\:{a}={sin}^{−} \left({x}\right),{b}={sin}^{−} \left({y}\right),{sin}^{−} \left({z}\right)={c} \\ $$$${sin}\left({a}\right)={x},{cos}\left({a}\right)=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }......\Rightarrow \\ $$$$\mathrm{1}\Leftrightarrow{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{y}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }+{z}\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }=\mathrm{2}{xyz} \\ $$$$ \\ $$