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) gt0`
D. `cos alpha lt 0`

Explanation

We observe that `sqrt(3)/(2)gt6/11gt1/2 and 0le4/9le1/2`
`therefore sin^(-1)sqrt(3)/(2)gtsin^(-1)6/11gtsin^(-1)(1/2)`
and `cos^(-1)0gt^(-1)(4/9)gtcos^(-1)(1/2)`
`rarr (pi)/(3)gtsin^(-1)(6/11)gt(pi)/(6)and (pi)/(2)gtcos^(-1)(4/9)gt(pi)/(3)`
`rarr pigtalphagt(pi)/(2)and (3pi)/(2)gtbetapi`
`rarr (pi)/(2)ltalpha,alphale(3pi)/(1)rarr(3pi)/(2)rarr(3pi)/(2)ltalpha+betalt(5pi)/(2)`
`rarr cos alphalt0 sin beta lt 0 cos beta lt 0`
Hence option b,c and d are correct