The minimum value of atan 2 x + cot 2 x equals the maximum value of asin 2 y + bcos 2 y where a and b are grater than 0. Then a/b is (a)2 (b)4 (c)6 (d)8 Please explain fully.

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Vikas TU · Answer

Minimum value of atan 2 x + cot 2 x = > First convert them into sec and cosec. = a(sec^2x – 1) + (cosec^2 – 1) = a(1/cos^^2x – 1) + (1/sin^2x – 1) = 1/sin^2x + a/cos^2x – 2 = 1/(1-cos^2x) + a/cos^2x – 2 Now here sin and cos values should be maximum for getting minmum values. SO, Min. value would be = > 1/(1-0) + a/1 – 2 => a – 1 Now maximum value of asin 2 y + bcos 2 y = > asin^2y +b – bsin^2y => b + [sin^2y]*(a – b) => b + a – b => a

Mudit · Answer

The minimum value is 2&radic;ab but I can not understand how it came?

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The minimum value of atan 2 x + cot 2 x equals the maximum value of asin 2 y + bcos 2 y where a and b are grater than 0. Then a/b is (a)2 (b)4 (c)6 (d)8 Please explain fully.

The minimum value of atan²x + cot²x equals the maximum value of asin²y + bcos²y where a and b are grater than 0. Then a/b is
(a)2 (b)4 (c)6 (d)8
Please explain fully.

2 Answers

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The minimum value of atan 2 x + cot 2 x equals the maximum value of asin 2 y + bcos 2 y where a and b are grater than 0. Then a/b is (a)2 (b)4 (c)6 (d)8 Please explain fully.

The minimum value of atan2x + cot2x equals the maximum value of asin2y + bcos2y where a and b are grater than 0. Then a/b is(a)2 (b)4 (c)6 (d)8Please explain fully.

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The minimum value of atan²x + cot²x equals the maximum value of asin²y + bcos²y where a and b are grater than 0. Then a/b is
(a)2 (b)4 (c)6 (d)8
Please explain fully.