tan-1(x) is inverse tan or arctan and is not the same as 1/tan(x). (tan(x))-1=1/tan(x)=cot(x). -1 is not an exponent when applied to a function (sin-1, cos-1, etc.)--it is used in mathematics to represent an inverse function.
I suspect that the question is meant to be:
Prove tan-1(½tan(2A))+tan-1(cot(A))+tan-1((cot(A))3)=0.
Let tan(a)=½tan(2A), tan(b)=cot(A), tan(c)=cot3(A), then
a+b+c=0 if the identity is true.
Also tan(a+b+c)=0 if the identity is true, and it is this form of the equation that's used here to prove the identity.
tan(a+b+c)=tan((a+b)+c)=(tan(a+b)+tan(c))/(1-tan(a+b)tan(c)).
tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))=
(½tan(2A)+cot(A))/(1-½tan(2A)cot(A)).
tan(2A)=2tan(A)/(1-tan2(A)), so ½tan(2A)cot(A)=1/(1-tan2(A)) (because cot(A)tan(A)=1) and:
tan(a+b)=(tan(A)/(1-tan2(A))+cot(A))/(1-1/(1-tan2(A))), multiply top and bottom by 1-tan2(A):
tan(a+b)=-(tan(A)+cot(A)(1-tan2(A)))/tan2(A)=-(tan(A)+cot(A)-tan(A))/tan2(A)=-cot(A)cot2(A)=-cot3(A)
tan(a+b+c)=(-cot3(A)+cot3(A))/(1+cot6(A))=0.
Therefore, tan-1(½tan(2A))+tan-1(cot(A))+tan-1((cot(A))3)=0 QED