If A=[1tan x−tan x1],ATA−1=
[cos 2x−sin 2xsin 2xcos 2x]
A=[1tan x−tan x1]|A|=1+tan2 xadj(A)=[1−tan xtan x1]A−1=11+tan2 x[1−tan xtan x1]AT=[1−tan xtan x1]∴ATA−1=11+tan2 x[1−tan xtan x1][1−tan xtan x1]=11+tan2 x[1−tan2x−2 tan x2 tan x1−tan2x]=⎡⎢⎣1−tan2x1+tan2x−2 tan x1+tan2 x2 tan x1+tan2 x1−tan2 x1+tan2x⎤⎥⎦=[cos 2x−sin 2xsin 2xcos 2x]