In a triangle A B C , tan A 2 tan B 2 tan C 2 2 ≤

In a triangle ABC,tanA2tanB2tanC22
  1. 127
  2. 118
  3. 19
  4. 13

Solution

We know that for ABC

tanA2tanB2+tanB2tanC2+tanC2tanA2=1   ....i

For triangle, tanA2, tanB2, tanC2>0

Now,

AMGM

tanA2tanB2+tanB2tanC2+tanC2tanA23tanA2tanB2×tanB2tanC2×tanC2tanA213

tan2A2tan2B2tan2C21313

tanA2tanB2tanC22127

Asked in: AP EAMCET 2022 (04 Jul Shift 2)

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