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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.
Minimum value of atan2x + cot2x = >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 – 2Now 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 asin2y + bcos2y = > asin^2y +b – bsin^2y=> b + [sin^2y]*(a – b)=> b + a – b=> a
The minimum value is 2√ab but I can not understand how it came?
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