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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) gt0`D. `cos alpha lt 0` |
Answer» 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 |
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