Chapter 2 Inverse Trigonometric Functions
- cos-1(\(\frac { 1 }{ 2 }\)) + 2 sin-1(\(\frac { 1 }{ 2 }\)) cos-1( 1 / 2 ) + 2 sin-1( 1 / 2 ) Now, cos-1( 1 / 2 ) = π / 3 and sin-1( 1 / 2 ) = π / 6 ∴ cos-1( 1 / 2 ) + 2 sin-1( 1 / 2 ) = π / 3 + 2 x π... 1
- If sin-1 = y, then (A) 0 ≤ y ≤ π (B) – \(\frac { π }{ 2 }\) ≤ y ≤ \(\frac { π }{ 2 }\) (C) 0 < y < π optin B The range of principal value branch of sin-1 is [- π / 2 , π / 2 ]∴ if sin-1 x = y, then – π / 2 ≤ y ≤ π / 2 1
- tan-1\(\sqrt{3}\) – sec-1(-2) is equal to (A) π (B) – \(\frac { π }{ 3 }\) (C) \(\frac { π }{ 3 }\) option B tan-1 √(3) = π / 3 and sec-1(-2) = π – π / 3 = 2π / 3 , since principal value branch of sec-1 is [0, π] – π / 2 ∴ tan-1 √(3) – sec-... 1