If  α=3sin−1611 and β=3cos−149,  where the inverse trigonometric functions take only the principal values, then the correct options(s) is (are)

α=3sin−1611,611>612⇒sin-1611>sin-112⇒α3>π6⇒α>π2⇒cosα<0,sinα>0β=3cos−149,49<48⇒cos-149>cos-112⇒β3>π3⇒β>π⇒sinβ<0,cosβ<0Now, cosα+β=cosαcosβ−sinαsinβ>0