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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.

Grade:12

2 Answers

Vikas TU
14149 Points
6 years ago
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 –  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 asin2y + bcos2 = > asin^2y +b – bsin^2y
=> b + [sin^2y]*(a – b)
=> b + a – b
=> a
Mudit
67 Points
6 years ago
The minimum value is 2√ab but I can not understand how it came?
 

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