But is it a trig question...

Let A be a number such that:

sec(A^(A+1))=cosh(A^(A+1))

Let B be a number such that:

tan(B+ln(-1)/i-((pi)^1/2)/2)+sinh(B-((pi)^1/2)/2)=0

The real part of e^(i*pi((A-(B+1))/(2^(A-B)+3))) can be expressed as K^(1/2).

Find i^(K^(-1)).

Note: ln(-1) denotes all possible compex numbers such that e^x=-1, sinh x and cosh x have equations sinh x=(e^x-e^(-x))/2 and cosh x=(e^x+e^(-x))/2, and i denotes the number that satisfies x^2=-1.