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If A > 0, B > 0, and A + B = `pi/3` then the maximum value of tan A tan B is |
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Answer» Correct Answer - 3 `A+b= pi//3` `rArr tan (A+B) = sqrt3` `rArr (tanA + tanB)/(1-tan A tan B) = sqrt3` `rArr = (tan A + (y)/(tan A)) /( 1-y) = sqrt3` ` rArr tan^(2)A+ sqrt3 (y-1) tan A +y =0 ` For real values of `tan A`, `" " 3(y-1)^(2) - 4y =0` `rArr 3y^(2) - 10y + 3 ge 0` `rArr (y-3)(3y-1) ge 0` `rArr y le (1)/(3) or y ge 3` But A, B `gt` 0 and A +B `= pi//3` `rArr A, B lt pi//3` `rArr tan A tan B lt 3` so, ` ge 3` is not a possibility. Therefore, `y le (1)/(3)` i.e., max. value of `y ` is `1//3` |
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