If m andx are two real numbers, then the real value of e2micot−1xxi+1xi−1m is equal to

Letcot−1x=θ⇒cotθ=x and tanθ=1x Now,e2icot−1x=e2iθ=cos2θ+isin2θ                                    =1−tan2θ1+tan2θ+i2tanθ1+tan2θ                                   =x2−1x2+1+i2xx2+1=(x+i)2(x−i)(x+i)                                 =x+ix−i =ix−1ix+1⇒ix+1ix−1e2icot−1x=1Raising both sides to the power m, we get the real value of e2micot−1x(ix+1ix−1)m=1