Question

证明反三角函数公式arctanA-arctanB=arctan(A-B)/1+AB，arctanA+arctanB=ar

Answer 1

a=arctanA b=arctanB tan(a)=A tan(b)=B

tan(a-b)=tan(a)-tan(b)/(1+tan(a)tan(b))=(A-B)/(1+AB)

a-b=arctan((A-B)/(1+AB))

arctanA-arctanB=arctan((A-B)/(1+AB))

同理tan(a+b)=tan(a)+tan(b)/(1-tan(a)tan(b))=(A+B)/(1-AB)

a+b=arctan((A+B)/(1-AB))

arctanA+arctanB=arctan((A+B)/(1-AB))