Prove that `tan^2x+6lns e c x+2cosx+4>6secx` for `x\ in ((3pi)/2,\ 2pi)dot`

Prove that tan^(2)x+6ln sec x+2cos x+4>6sec x for x in(3(pi)/(2),2 pi)

Let f (x) = 7 tan ^(8) x + 7 tan ^(6) x - 3 tan ^(4) x - 3 tan ^(2) x for all x in (-(pi)/(2), (pi)/(2)). Then the correct expression(s) is (are).

If f (x) = sqrt(cos ec ^(2) x - 2 sin x cos x - (1)/(tan ^(2) x )) x in ((7pi)/(4), 2pi ) then f' ((11 pi)/(6))=

Minimum value of f(x)=cos^(2)x+(secx)/(4) , x in (-(pi)/(2),(pi)/(2)) is

Prove that tan^(-1)((cos x)/(1+sin x))=(pi)/(4)-(x)/(2),|x in(-(pi)/(2),(pi)/(2))

Prove that tan^(-1)((cosx)/(1+sinx))=(pi/4-x/2)

Prove that (i) " tan"^(2) .(pi)/(3) + 2cos^(2) .(pi)/(4)+ 3 sec^(2).(pi)/(6)+ 4 cos^(2).(pi)/(2)=8 (ii) " sin ".(pi)/(6) " cos 0 + sin ".(pi)/(4) " cos " .(pi)(4) + " sin " .(pi)/(3) "cos " .(pi)/(6) =(7)/(4) (iii) " 4sin " (pi)/(6) " sin"^(2) (pi)/(3) + 3 " cos " .(pi)/(3) " tan ".(pi)/(4) = " cosec"^(2).(pi)/(2)=4

If 0

The area bounded by the curve y=cosx and y=sin 2x, AA x in [(pi)/(6), (pi)/(2)] is equal to