cos−1(−12)−2sin−1(12)+3cos−1(−1√2)−4tan−1(−1) equals to
19π/12
35π/12
47π/12
43π/12
cos−1(−12)−2sin−1(12)+3cos−1(−1√2)−4tan−1(−1) =cos−1(cos2π3)−2sin−1(sinπ6)+3cos−1(cos3π4)−4tan−1tan(−π4) =2π3−2.π6+3.3π4+4.π4 =2π3−π3+9π4+π =8π−4π+27π+12π12=43π12