Mathematics › Trigonometric Equations › Solving Trigonometric Inequalities
We know that for ∆ABCtanA2tanB2+tanB2tanC2+tanC2tanA2=1 ....iFor triangle, tanA2, tanB2, tanC2>0Now,AM≥GM⇒tanA2tanB2+tanB2tanC2+tanC2tanA23≥tanA2tanB2×tanB2tanC2×tanC2tanA213⇒tan2A2tan2B2tan2C213≤13⇒tanA2tanB2tanC22≤127
We know that for ∆ABC
tanA2tanB2+tanB2tanC2+tanC2tanA2=1 ....i
For triangle, tanA2, tanB2, tanC2>0
Now,
AM≥GM
⇒tanA2tanB2+tanB2tanC2+tanC2tanA23≥tanA2tanB2×tanB2tanC2×tanC2tanA213
⇒tan2A2tan2B2tan2C213≤13
⇒tanA2tanB2tanC22≤127
Asked in: AP EAMCET 2022 (04 Jul Shift 2)
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