Doubt

1. cos x can be neg. in 3rd quadrant but it can never be <1 . cos x=1/sec x = 1/(-0.7) i.e sumthing less than -1 for sure. Thus i think sec x cant have value of -0.7

2. Im not sure what the 1st part of the q is.I think it is |sin a|. Then the q. is |sin a|-cos B max. value. Let Sin A take its max. value of 1. Cos B take its min. value of -1. Then it becomes 1- -1,i.e 1+1=2.
Therefore max. value of expression is 2.

sec²x=1/cos²x.
(1/cos²x +cos²x)/2 >= √(1/cos²x * cos²x) Since AM >= GM
Therefore it comes--sec²x + cos²x >= 2.

Another way of solving question no 3 ....which is a bit long but easier to understand...

cos²Î¸ + sec²Î¸
=cos²Î¸ + sec²Î¸ +2secθcosθ - 2secθcosθ
=(cosθ - secθ)²+2secθcosθ
=(cosθ - secθ)²+2

since,
(cosθ - secθ)²â‰¥ 0
therefore (cosθ - secθ)² +2 ≥ 2
i,e cos²Î¸ + sec²Î¸â‰¥2

hence proved....

ronald's answer is much more logical for question no. 3. the concept of am>=gm is very impt from iit point of view......

Thanks for ur answers.I had also found my technique for q no 3.

Since we are required 2 find minimum value of the expression and we have a '+' sign in between hence the minimum value of the expression has to b the minimum values of the individual functions.thus minimum value of secθ is -1 and that of cosθ is -1.Thus squaring and adding the minimum value of the expression must be =2

1)secθ=-0.7 .cosθ=-10/7.therefore no quadrant False
3)sec2 θ +cos2 θ= 1+cos^4 θ=cos has a maximum value of 1 therefore it has a maximum value of 2 true