Question

tan(a+b)=1/2tan(a-b)=1/3.tan2a=______tan2b=_____详细过程,

Answer 1

解tan2a

=tan[(a+b）+（a-b）]

=[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]

=[1/2+1/3]/[1-1/2*1/3]

=[3+2]/(6-1)

=1

同理

tan2b

=tan[(a+b）-（a-b）]

=[tan(a+b)-tan(a-b)]/[1+tan(a+b)tan(a-b)]

=[1/2-1/3]/[1+1/2*1/3]

=[3-2]/(6+1)

=1/7.