Question

Question 14: Find the value of tan¯¹ √3 -sec¯¹ (-2) is equal to (A) π (B) -π/3 (C) π/3 (D) 2π/3

Class 12 - Math - Inverse Trigonometric Functions

Answer 1

Let −1√3 = , then tan = √3 = /3

We know that the range of the principal value branch of tan−1 is (- π /2 , π /2 ).
∴ tan-1√3 = π /3
Let sec-1 (-2) = y, then
sec y = -2 = -sec π /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
So, the option (B) is correct.