The minimum value of $a t a n^{2} x + b c o t^{2} x$ equals the maximum value of a $s i n^{2} θ + b c o s^{2} θ w h e r e a > b > 0,$ then

a = b

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a = 2b

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a = 3b

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a = 4b

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Solution

The correct option is D
a = 4b

Minimum value of a $t a n^{2} x + b c o t^{2} x i s 2 \sqrt{a b}$ and maximum value of a $s i n^{2} θ + b c o s^{2} θ$ is a. $∵ a s i n^{2} θ + b c o s^{2} θ = (a - b s i n^{2} θ + b)$
Given, $a = 2 \sqrt{a b}$ Hence (d) is the correct answer.

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