Mathematics - Properties of Triangle Question with Solution | TestHub

This is a standard identity in properties of triangles. We know that $r=4R\sin (A/2)\sin (B/2)\sin (C/2)$, $r_1=4R\sin (A/2)\cos (B/2)\cos (C/2)$, $r_2=4R\cos (A/2)\sin (B/2)\cos (C/2)$, and $r_3=4R\cos (A/2)\cos (B/2)\sin (C/2)$. Summing these, $r_1+r_2+r_3-r=4R[\sin (A/2)\cos (B/2)\cos (C/2)+\cos (A/2)\sin (B/2)\cos (C/2)+\cos (A/2)\cos (B/2)\sin (C/2)-\sin (A/2)\sin (B/2)\sin (C/2)]$. Let $A/2=x,B/2=y,C/2=z$. Then $x+y+z=A/2+B/2+C/2=(A+B+C)/2=180^{\circ }/2=90^{\circ }$. The expression in the bracket is $\sin x\cos y\cos z+\cos x\sin y\cos z+\cos x\cos y\sin z-\sin x\sin y\sin z$, which is the expansion of $\sin (x+y+z)$. Since $x+y+z=90^{\circ }$, $\sin (x+y+z)=\sin 90^{\circ }=1$. Therefore, $r_1+r_2+r_3-r=4R(1)=4R$.