If α = 3 sin - 1 ⁡ 6 11 a n d β = 3 cos - 1 ⁡ 4 9 , where the inverse trigonometric functions take only the principal values, then the correct options(s) is(are)

Step-by-step solution

α = 3 sin - 1 ⁡ 6 11 Since π 4 > sin - 1 ⁡ 6 11 > π 6 3 π 4 > 3 sin - 1 ⁡ 6 11 > π 2 cos ⁡ α < 0 cos ⁡ 3 cos - 1 ⁡ 4 9 = 4 × 4 9 3 - 3 × 4 9 = 4 × 64 729 - 12 9 = 256 - 12 × 81 72 < 0 5 π 12 > cos - 1 ⁡ 4 9 > π 3 5 π 4 > 3 cos - 1 ⁡ 4 9 > π sin ⁡ 3 c o s - 1 4 9 < 0 3 π 4 > 3 sin - 1 ⁡ 6 11 > π 2 5 π 4 > 3 cos - 1 ⁡ 4 9 > π 2 π > 3 sin - 1 ⁡ 6 11 + 3 cos - 1 ⁡ 4 9 > 3 π 2 cos ⁡ 3 sin - 1 ⁡ 6 11 + 3 cos - 1 ⁡ 4 9 > 0