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In any triangle ABC `sum (sin^2A+sinA+1)/sinA` is always greater than or equal |
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Answer» In triangle ABC, `sum (sin^2A + sin A+1)/sinA = (sin^2A + sin A+1)/sinA+ (sin^2B + sin B+1)/sinB+ (sin^2C + sin C+1)/sinC` `=sinA+1/sinA+sinB+1/sinB+sinC+1/sinC+3` As, `sin theta` in any triangle is always greater than `0`. `:. sin theta + 1/sin theta ge 2`. `:. sinA+1/sinA+sinB+1/sinB+sinC+1/sinC ge6 ``=>sinA+1/sinA+sinB+1/sinB+sinC+1/sinC+3 ge 9` Thus, `sum (sin^2A + sin A+1)/sinA ge 9`. |
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