If A=1tan⁡x−tan⁡x1, then ATA-1 is

|A|=1tan⁡x−tan⁡x1=1+tan2⁡x≠0So A is invertible. Also,adj⁡A=1tan⁡x−tan⁡x1T=1−tan⁡xtan⁡x1Now, A−1=1|A|adj⁡A               =11+tan2⁡x1−tan⁡xtan⁡x1=11+tan2⁡x−tan⁡x1+tan2⁡xtan⁡x1+tan2⁡x11+tan2⁡x∴ ATA−1=1−tan⁡xtan⁡x111+tan2⁡x−tan⁡x1+tan2⁡xtan⁡x1+tan2⁡x11+tan2⁡x=1−tan2⁡x1+tan2⁡x−2tan⁡x1+tan2⁡x2tan⁡x1+tan2⁡x1−tan2⁡x1+tan2⁡x=cos⁡2x−sin⁡2xsin⁡2xcos⁡2x