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.

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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.