Question

tan-1\(\sqrt{3}\) – sec-1(-2) is equal to

(A) π

(B) – \(\frac { π }{ 3 }\)

(C) \(\frac { π }{ 3 }\)

(D) \(\frac { 2π }{ 3 }\)

Answer 1

option B

tan-1\(\sqrt{3}\) = \(\frac { π }{ 3 }\)

and sec-1(-2) = π – \(\frac { π }{ 3 }\) = \(\frac { 2π }{ 3 }\),

since principal value branch of sec-1 is [0, π] – {\(\frac { π }{ 2 }\)}

∴ tan-1\(\sqrt{3}\) – sec-1(-2) = \(\frac { π }{ 3 }\) – \(\frac { 2π }{ 3 }\)

= – \(\frac { π }{ 3 }\)