Question

the minimum value of 2÷3 tan^2 theta+ 27÷2 cot^2 theta​

Answer 1

We have to find the minimum value of 2/3 tan²θ + 27/2 cot²θ

Solution : 2/3 tan²θ + 27/2 cot²θ

here terms (2/3)tan²θ and (27/2)cot²θ both are positive.

so we can apply AM ≥ GM

AM = arithmetic mean of given terms

GM = geometric mean of given terms

AM = {(2/3)tan²θ + (27/2)cot²θ}/2

GM = {(2/3)tan²θ × (27/2)cot²θ}½ = {(2/3) × (27/2) tan²θcot²θ}½ = 3

now, {(2/3)tan²θ + (27/2)cot²θ}/2 ≥ 3

⇒2/3 tan²θ + 27/2 cot²θ ≥ 6

Therefore the minimum value of 2/3 tan²θ + 27/2 cot²θ is 6.

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Answer 2

Answer:

The minimum value of 2/3 tan²θ + 27/2 cot²θ is 6.