[" The solution set of Tan "(4k+2)x-Tan(4k+],[" 1) "x-" Tan "(4k+2)x" Tan "(4k+1)x=1,k in Z" is "]

The solution set tan(4k+2)x-tan(4k+1)x-tan(4k+2)x tan(4k+1)x=1,k in Z is

tan 4x = (4 tan x (1- tan ^(2) x ))/( 1 - 6 tan ^(2) x + tan ^(4) x)

tan 4x = (4 tan x (1- tan ^(2) x ))/( 1 - 6 tan ^(2) x + tan ^(4) x)

If tan x-tan^(2)x=1, then the value of tan^(4)x-2tan^(3)x-tan^(2)x+2tan x+k=4, then k is equal to

If int(cos 4x + 1)/(cot x - tan x)dx = k cos 4x + c , then k is

The value of k + f(0) so that f (x)={{:((sin (4k-1)x)/(3x)"," , x lt 0),((tan (4k +1)x )/(5x)"," , 0 lt x lt (pi)/(2)), ( 1"," , x =0):} can bemade continous at x=0 is:

The value of k + f(0) so that f (x)={{:((sin (4k-1)x)/(3x)"," , x lt 0),((tan (4k +1)x )/(5x)"," , 0 lt x lt (pi)/(2)), ( 1"," , x =0):} can bemade continous at x=0 is:

Prove that: tan4x=(4tan x(1-tan^(2)x))/(1-6tan^(2)x+tan^(4)x)

Prove the following: tan 4x =(4tan x (1-tan^2x))/(1-6 tan^2x + tan^4x)