1. **Problem statement:** Show that $$\frac{2b}{c+a-b} = \frac{\cot \frac{C}{2} + \cot \frac{A}{2}}{\cot \frac{B}{2}}$$ for triangle ABC with sides $a,b,c$ opposite angles $A,B,C$ respectively. 2. **Relevant formulas:** - Half-angle cotangent formulas in a triangle: $$\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$$ $$\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$$ $$\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$$ where $s = \frac{a+b+c}{2}$ is the semiperimeter. 3. **Express the right side:** $$\frac{\cot \frac{C}{2} + \cot \frac{A}{2}}{\cot \frac{B}{2}} = \frac{\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} + \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}}{\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}$$ 4. **Simplify numerator:** Find common denominator inside numerator: $$= \frac{\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \cdot \sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}} + \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \cdot \sqrt{\frac{(s-a)(s-b)}{(s-a)(s-b)}}}{\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}$$ $$= \frac{\sqrt{\frac{s(s-c)(s-b)(s-c)}{(s-a)(s-b)(s-b)(s-c)}} + \sqrt{\frac{s(s-a)(s-a)(s-b)}{(s-b)(s-c)(s-a)(s-b)}}}{\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}$$ 5. **Simplify inside roots:** $$= \frac{\sqrt{\frac{s(s-c)^2 (s-b)}{(s-a)(s-b)^2 (s-c)}} + \sqrt{\frac{s(s-a)^2 (s-b)}{(s-b)^2 (s-c)(s-a)}}}{\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}$$ 6. **Cancel terms:** $$= \frac{\sqrt{\frac{s(s-c)^2 (s-b)}{(s-a)(s-b)^2 (s-c)}} + \sqrt{\frac{s(s-a)^2 (s-b)}{(s-b)^2 (s-c)(s-a)}}}{\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}$$ 7. **Rewrite numerator terms:** $$= \frac{\frac{(s-c)}{(s-b)} \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} + \frac{(s-a)}{(s-b)} \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}{\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}$$ 8. **Factor numerator:** $$= \frac{\left(\frac{(s-c)+(s-a)}{(s-b)}\right) \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}{\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}}$$ 9. **Cancel the square root terms:** $$= \frac{(s-c)+(s-a)}{(s-b)} = \frac{2s - c - b}{s - b}$$ 10. **Recall $s = \frac{a+b+c}{2}$:** $$2s = a + b + c$$ So numerator: $$2s - c - b = a + b + c - c - b = a$$ 11. **Therefore:** $$\frac{\cot \frac{C}{2} + \cot \frac{A}{2}}{\cot \frac{B}{2}} = \frac{a}{s - b}$$ 12. **Compare with left side:** Given left side is $$\frac{2b}{c + a - b}$$ Rewrite denominator: $$c + a - b = (a + b + c) - 2b = 2s - 2b = 2(s - b)$$ 13. **So left side becomes:** $$\frac{2b}{2(s - b)} = \frac{b}{s - b}$$ 14. **Conclusion:** The left side is $$\frac{b}{s - b}$$ and the right side is $$\frac{a}{s - b}$$, so the original equality as stated is not generally true unless $a = b$. **Note:** The problem as stated seems to have a discrepancy. Possibly a typo or misinterpretation in the problem statement. **Slug:** triangle half-angle cot **Subject:** geometry