1.

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

Answer» Correct Answer - B::C::D
Here, `alpha = 3sin ^(-1) ((6)/(11)) and beta = 3 cos ^(-1) ((4)/(9)) "as " (6)/(11) gt (1)/(2)`
`rArr sin^(-1)((6)/(11)) gt sin ^(-1)((1)/(2)) = (pi)/(6)`
`therefore " " alpha = 3 sin ^(-1) ((6)/(11)) gt (pi)/(2)`
`rArr " " cos alpha lt 0`
Now, `" " beta = 3 cos ^(-1)((4)/(9))`
As ` " " (4)/(9) lt (1)/(2) rArr cos ^(-1) ((4)/(9)) gt cos ^(-1) ((1)/(2)) = (pi)/(3)`
`therefore " " beta = 3 cos ^(-1) ((4)/(9)) gt pi`
`therefore cos beta lt 0 and sin beta lt 0`
Now, `alpha + beta ` is slightly greater that ` ( 3pi )/(2)`.
`therefore " " cos ( alpha + beta ) gt 0`


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