If `0lt=xlt=pi/3` then range of `f(x)=sec(pi/6-x)+sec(pi/6+x)` is

Find the range of f(x)=sec(pi/4cos^2x)

If - pi/2 lt x lt pi/2 , then the range of the function f(x)=cos [x] is

If 0lt=xlt=2pi, then the number of solutions of 3(sinx+cosx)-2(sin^3x+cos^3x)=8 is

Column I, Column II Range of f(x)=sin^(-1)x+cos^(-1)x+cot^(-1)xi s , p. [0,pi/2]uu[pi/2,pi] Range of f(x)= cot^(-1)x+tan^(-1)x+cos e c^(-1)xi s , q. [pi/2,(3pi)/2] Range of f(x)=cot^(-1)x+tan^(-1)x+cos^(-1)xi s , r. [0,pi] Range of f(x)= sec^(-1)x+cos e c^(-1)x+sin^(-1)xi s , s. [(3pi)/4,(5pi)/4]

Find from the first principle,the derivative of f(x)=sec2xat x=pi/8 .

If cos(pi/7), cos((3pi)/7), cos((5pi)/7) are the roots of the equation 8x^3-4x^2-4x+1=0 The value of sec(pi/7)+sec((3pi)/7)+sec((5pi)/7) is

Find from the first principle, the derivative of f(x)=sec2x" at "x=(pi)/(8)

Find the range of the function f(x) = (sec^2x - tan x)/(sec^2 x + tan x) , (-pi/2 lt x lt pi/2) .

The maximum value of the function f(x)=sin(x+pi/6)+cos(x+pi/6) in the interval (0,pi/2) occurs at (a) pi/(12) (b) pi/6 (c) pi/4 (d) pi/3

Let f(x) =(1- tanx)/(4x-pi) when 0 le x lt pi/2 and x ne pi/4, If f (x) is defined at x =pi/4, then the value of f ((pi)/(4)) is-