If `alpha=3sin^-1(6/11) `and `beta=3cos^-1(4/9)`, where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)

Options

A. `cos beta gt 0`
B. `sin beta lt 0`
C. `cos (alpha + beta) gt 0`
D. `cos alpha lt 0`

Explanation

`alpha = 3 sin^(-1).(6)/(11) gt 3 sin^(-1).(6)/(12) = 3 sin^(-1).(1)/(2) = (pi)/(2)`
and `beta = 3 cos^(-1).(4)/(9) gt 3 cos^(-1).(4)/(8) = pi`
So, `alpha gt (pi)/(2) and beta gt pi`
`:. alpha + beta gt (3pi)/(2)`