Let ABC be an isosceles triangle with base BC. If r is the radius of the circle inscribed in

$\mathrm{△}ABC\text{ and }{r}_1$ is the radius of the circle escribed opposite to the $\mathrm{\angle }A,$ then the product $r_1\cdot r$ can be equal to

$\because r=\frac{\mathrm{\Delta }}{S},r_1=\frac{\mathrm{\Delta }}{S-a}$

Now,   $r_1\cdot r=\frac{{\mathrm{\Delta }}^2}{s(s-a)}=\frac{s(s-a)(s-b)(s-c)}{s(s-a)}$ 

                  $\begin{array}{l}=(s-b)(s-c)=(s-b{)}^2\\ =\frac{(2s-2b{)}^2}{4}=\frac{(a+b+c-2b{)}^2}{4}\\ =\frac{a^4}{4} \text{ [option (d)] }\\ =\frac{4R^2{\sin }^{2}A}{4}=R^2{\sin }^{2}A\text{ [option (a)] }\end{array}$

Also, if          $\mathrm{\angle }B=\theta$

$\begin{array}{l}\Rightarrow \mathrm{\angle }A=\pi -2\theta \\ \therefore r_1\cdot r=R^2{\sin }^2(\pi -2\theta )\\ =R^2{\sin }^{2}2\theta \\ =R^2{\sin }^{2}2B \text{ [option (b)] }\end{array}$