Find the range of `f(x)=sec(pi/4cos^2x),` where `-oo lt x lt oo` is

Find the range of f(x)=cos^2x+sec^2x .

Find the intervals in which function f(x) = sin x-cos x, 0 lt x lt 2pi is (i) increasing, (ii) decreasing.

The primitive of the function f(x)= x | cos x| , when pi/2 lt x lt pi is given by

If 0lt=xlt=pi/3 then range of f(x)=sec(pi/6-x)+sec(pi/6+x) is (a)(4/(sqrt(3)),oo) (b) (4/(sqrt(3)),oo) (c)(0,4/(sqrt(3))) (d) (0,4/(sqrt(3)))

In the interval ((pi)/(4), (11 pi)/(12)) , then function f (x) = sin 3x - cos 3x, 0 lt x lt pi is:

If exp [(sin^(2)x+sin^(4) x +sin^(6)x+.... oo)" In" 2] satisfies the equation y^(2)-9y+8=0 , then the value of (cos x )/( cos x+ sin x ),0 lt x lt (pi)/(2) , is

If sec x cos 5x+1=0 , where 0lt x lt 2pi , then x=

Find the local maxima and local minima of the functions: (i) f(x) = (sin x - cos x), " When " 0 lt x lt (pi)/(2) (ii) f(x) = (2 cos x + x), " when " 0 lt x lt pi