Question

tan⁻¹√3−sec⁻¹(−2) is equal to

(A) π                             (B) − π/3
(C) π /3                         (D) 2 π/3

Answer 1

∴tan^-1√3=π/3
Let sec^-1(-2)=y, then

 sec y=-2=-secπ/3=sec(π-π/3)=sec(2π/3)
We know that the range of the principal value branch of sec⁻¹ is [0,π]-{π/2}

∴sec^-1(-2)=2π/3

Now, 

tan^-1√3-sec^-1(-2)=π/3-2π/3=-π/3

Hence, the option (B) is correct.