$$ABC$$ is a right triangle in which $$\angle B = 90^\circ$$. The inradius of the triangle is $$r$$ and the circumradius of the triangle is $$R$$. If $$R \colon r = 5 \colon 2$$, then the value of $$\cot^2 \frac{A}{2} + \cot^2 \frac{C}{2}$$ is
Solution
Using $$\frac{r}{R} = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$ with $$B = 90^\circ$$ and $$\frac{r}{R} = \frac{2}{5}$$, solving for $$A$$ (with $$C = 90^\circ - A$$) gives a specific angle, and substituting back yields $$\cot^2\frac{A}{2} + \cot^2\frac{C}{2} = 13$$.
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