已知tan-1/1+tana=^2.求cot(a+4/π)

(tana-1)/(1+tana)=√2(tana+tan3π/4)/(1-tana*tan3π/4)=√2所以tan(a+3π/4)=√2tan的周期是π所以tan(a+π/4)=tan(a+3π/4-π)=√2所以cot(a+π/4)=1/tan(a+π/4)=1/√2=√2/2