Trigonometry 1

As $0<A,B,C<\pi$ , $\sin A,\sin B,\sin C>0$

$1=\cos A\cos B+\sin A\sin B\sin^2C\ge\cos A\cos B+\sin A\sin B=\cos(A-B)$

But $\cos(A-B)\le1$

So, we need $\cos(A-B)=1\implies A-B=2m \pi$

But $0<A,B<\pi \implies A-B=0$ and consequently $\sin^2C=1\implies C=\frac{\pi}{2}$

Now use $r=R(\cos A+\cos B+\cos C-1)=R(\sqrt2-1)$

$Put\quad C={ 90 }^{ o }\quad A=B={ 45 }^{ 0 }\quad \quad \quad (Hit\quad and\quad trial\quad Method)\\ Which\quad satisfy\quad the\quad equation\quad cosAcosB+sinAsinB{ (sinC) }^{ 2 }=1\\ now\quad take\quad a=b=1\quad and\quad c=\sqrt { 2 } \\ r=\frac { \Delta }{ s } =\frac { area }{ semiperimeter } =\frac { (\frac { 1 }{ 2 } *1*1) }{ (\frac { 1+1+\sqrt { 2 } }{ 2 } ) } =\frac { 1 }{ 2+\sqrt { 2 } } \\ R=\frac { a.b.c }{ 4\Delta } =\frac { 1.1.\sqrt { 2 } }{ 4.\frac { 1.1.1 }{ 2 } } =\frac { 1 }{ \sqrt { 2 } } \\ \frac { r }{ R } =\frac { \frac { 1 }{ 2+\sqrt { 2 } } }{ \frac { 1 }{ \sqrt { 2 } } } =\frac { 1 }{ 1+\sqrt { 2 } } \approx 0.414213562$