If `x in (pi,(3pi)/2),`then `4cos^2(pi/4-x/2)+sqrt(4sin^4x+sin^2 2x)` is always equal to (a) 1 (b) 2 (c) `-2` (d) none of these

Explanation

` x in (pi,3/2pi)->sintheta=-ve`

`4(cos^2(pi/4-x/2))+sqrt(4sin^4x+4sin^2xcos^2x)`

`=2(2cos^2(pi/4-x/2))+sqrt(4sin^2x(sin^2x+cos^2x))`

`=2(1+cos(pi/2-x))+2|sinx|`

`=2+2sinx+(-2sinx)`

`=2`

Option B is correct.