For $\triangle ABC,$ $r_1+r_3+r=r_2$, find $\sec^2A+\csc^2B-\cot^2C.$ In a triangle $ABC,$ if $r_1+r_3+r=r_2$,then find the value of $\sec^2A+\csc^2B-\cot^2C.$,where symbols have their usual meanings.

Here $r_1=\dfrac{\Delta}{s-a},r_2=\dfrac{\Delta}{s-b},r_3=\dfrac{\Delta}{s-c},r=\dfrac{\Delta}{s}$
I put these values and simplified $r_1+r_3+r=r_2$ to get  $\dfrac{c^2}{\sin^2\dfrac{C}{2}}=\dfrac{b^2}{\cos^2\dfrac{B}{2}}$
I am stuck here. Please help.

$$r_1+r_3+r-r_2=0$$
We can write $$\frac{\Delta}{s-a}+\frac{\Delta}{s-c}+\frac{\Delta}{s}-\frac{\Delta}{s-b}=0$$
to have
$$\frac{1}{b+c-a}+\frac{1}{a+b-c}+\frac{1}{a+b+c}-\frac{1}{c+a-b}=0$$
and so
$$\frac{4b(a^2+c^2-b^2)}{(b+c-a)(a+b-c)(a+b+c)(c+a-b)}=0$$
Since we have $b^2=a^2+c^2$, we know that $\triangle{ABC}$ is a right triangle with $\angle B=90^\circ$.
Therefore, 
$$\sec^2A+\csc^2B-\cot^2C=\left(\frac bc\right)^2+1^2-\left(\frac ac\right)^2=\frac{(c^2+a^2)+c^2-a^2}{c^2}=\color{red}{2}.$$
