If 0≤x≤1, thentan⁡12sin−1⁡2x1+x2+12cos−1⁡1−x21+x2 is equal to

We know thatsin−1⁡2x1+x2=2tan−1⁡x, if −1≤x≤1and,    cos−1⁡1−x21+x2=2tan−1⁡x, if 0≤x<∞∴     tan⁡12sin−1⁡2x1+x2+12cos−1⁡1−x21+x2=    tan⁡2tan−1⁡x for 0≤x≤1=    tan⁡tan−1⁡2x1−x2=2x1−x2