Given `f(x) = tan(sqrt(pi^2/16 - x^2))` and A = R - [0,1]

STATEMENT-1: The values of f(x)=3sin(sqrt((pi^(2))/(16)-x^(2)) lie in the interval [0,(3)/(sqrt(3))]. and STATEMENT-2The domain of definition of the function f(x)=3sin(sqrt((pi^(2))/(16)-x^(2)) is [-(pi)/(4),(pi)/(4)]

If f (x) = tan ^(-1)sqrt((1 + sin x )/(1 - sin x)), 0 le x le (pi)/(2) then f' ((pi)/(6)) =?

If f (x) = tan ^(-1)sqrt((1 + sin x )/(1 - sin x)), 0 le x le (pi)/(2) then f' ((pi)/(6)) =?

Find the simplest value of f(x)=tan^(-1)((sqrt(1+x^(2))-1)/(x)),x in R-{0}

If f(x) = tan^(-1)(sqrt((1+sinx)/(1-sinx))), 0 lt x lt pi/2 , then f'(pi/6) is

If f(x) = tan^(-1)(sqrt((1+sinx)/(1-sinx))), 0 lt x lt pi/2 , then f'(pi/6) is

f(x)=2x-tan^(-1)x-ln(x+sqrt(1+x^(2)));x in R then

Let f(x)=2x-tan^-1 x- ln(x+sqrt(1+x^2)); x in R, then