Show that \[{r_1}{r_2}{r_3} = {r^3}{\cot ^2}\dfrac{A}{2}{\cot ^2}\dfrac{B}{2}{\cot ^2}\dfrac{C}{2}\] .

Hint: We now that in $ \vartriangle ABC $ , $ r $ is the radius of in circle and $ {r_1} $ , $ {r_2} $ and $ {r_3} $ are the radii of inscribed circle opposite to the vertices A, B and C respectively. We need to use the relations between the in radius and half angles as well as between the ex-radii and half angles to prove the given equation.
Formulas used:
If $ R $ is the radius of circum-circle of $ \vartriangle ABC $ , $ r $ is the radius of in circle and $ {r_1} $ , $ {r_2} $ and $ {r_3} $ are the radii of escribed circle opposite to the vertices A, B and C respectively, then
\[r = 4R\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\]
\[{r_1} = 4R\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)\]
\[{r_2} = 4R\cos \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)\]
\[{r_3} = 4R\cos \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\]

Complete step-by-step answer:
We are given that $ L.H.S. = {r_1}{r_2}{r_3} $
But, we know that \[{r_1} = 4R\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)\] , \[{r_2} = 4R\cos \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)\] and \[{r_3} = 4R\cos \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\]. Therefore we can say that
\[{r_1}{r_2}{r_3} = \;\;\left[ {4R\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)} \right] \times \left[ {4R\cos \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)} \right] \times \left[ {4R\cos \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)} \right]\]
Now, we will divide both the sides by $ {r^3} $ and putting the value \[r = 4R\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\] on the right side.
\[ \Rightarrow \dfrac{{{r_1}{r_2}{r_3}}}{{{r^3}}} = \dfrac{{\;\left[ {4R\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)} \right] \times \left[ {4R\cos \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)} \right] \times \left[ {4R\cos \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)} \right]}}{{{{\left[ {4R\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)} \right]}^3}}}\]
Simplifying on the right side, we get
\[
   \Rightarrow \dfrac{{{r_1}{r_2}{r_3}}}{{{r^3}}} = \dfrac{{64{R^3}\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right){{\cos }^2}\left( {\dfrac{A}{2}} \right){{\cos }^2}\left( {\dfrac{B}{2}} \right){{\cos }^2}\left( {\dfrac{C}{2}} \right)}}{{64{R^3}{{\sin }^3}\left( {\dfrac{A}{2}} \right){{\sin }^3}\left( {\dfrac{B}{2}} \right){{\sin }^3}\left( {\dfrac{C}{2}} \right)}} \\
   \Rightarrow \dfrac{{{r_1}{r_2}{r_3}}}{{{r^3}}} = \dfrac{{{{\cos }^2}\left( {\dfrac{A}{2}} \right){{\cos }^2}\left( {\dfrac{B}{2}} \right){{\cos }^2}\left( {\dfrac{C}{2}} \right)}}{{{{\sin }^2}\left( {\dfrac{A}{2}} \right){{\sin }^2}\left( {\dfrac{B}{2}} \right){{\sin }^2}\left( {\dfrac{C}{2}} \right)}} \\
 \]
We know that $ \dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta $
\[
   \Rightarrow \dfrac{{{r_1}{r_2}{r_3}}}{{{r^3}}} = {\cot ^2}\left( {\dfrac{A}{2}} \right){\cot ^2}\left( {\dfrac{B}{2}} \right){\cot ^2}\left( {\dfrac{C}{2}} \right) \\
   \Rightarrow {r_1}{r_2}{r_3} = {r^3}{\cot ^2}\left( {\dfrac{A}{2}} \right){\cot ^2}\left( {\dfrac{B}{2}} \right){\cot ^2}\left( {\dfrac{C}{2}} \right) = R.H.S. \\
 \]
Hence, it is proved that \[{r_1}{r_2}{r_3} = {r^3}{\cot ^2}\dfrac{A}{2}{\cot ^2}\dfrac{B}{2}{\cot ^2}\dfrac{C}{2}\]
So, the correct answer is “\[{r_1}{r_2}{r_3} = {r^3}{\cot ^2}\dfrac{A}{2}{\cot ^2}\dfrac{B}{2}{\cot ^2}\dfrac{C}{2}\]”.

Note: Here, we have seen the terms in circle and inscribed circles of triangle. Let us understand both the terms. When a circle is constructed inside the triangle in such a way that the circle touches all the sides of the triangle is called an in-circle or an inscribed circle. The centre of the incircle is known as an in centre of the triangle. The incentre is also the point of intersection of internal bisectors of the triangle. The radius of the in circle is called the in radius of triangle. The circle which touches one of the sides of the triangle is called an escribed circle. Therefore, an escribed circle is in the exterior of the triangle.