Given,
tan(2tan−151+sec−125+2tan−181)
=tan(2(tan−151+tan−181)+tan−1(21))
=tan(2(tan−11−40151+81)+tan−1(21))
=tan[2tan−1(31)+tan−1(21)]
=tan[tan−1(1−(31)22×31)+tan−1(21)]
=tan[tan−1(43)+tan−1(21)]
=tan[tan−11−8343+21]
=tantan−18545
=tantan−12
=2