Question

If in $△ABC,co{s}^{2}A+co{s}^{2}B+co{s}^{2}C=1,$ then the $△ABC$ is right angled.

Answer 1

From the given relation, we have
$co{s}^{2}A+co{s}^{2}B-(1-co{s}^{2}C)=0$
or $co{s}^{2}A+(co{s}^{2}B-si{n}^{2}C)=0$
or $co{s}^{2}A+cos(B+C)cos(B-C)=0$
or $cosA[-cos(B+C)-cos(B-C\left)\right]=0$
or $2cosAcosBcosC=0$
Hence either $cosA=0,cosB=0$ or $cosC=0$
i.e. either $A=\pi /2$ or $B=\pi /2$ or $C=\pi /2$.
It follows that the $△ABC$ is right angled.