" finit "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 f(x)=int(tan^(3)x-x tan^(2)x)dx Where f(0)=0 and lim_(x rarr0)(f(x))/(x^(m)) is non-zero finite then m=

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)

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

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

If lim_(x rarr0)((cos4x+a cos2x+b)/(x^(4))) is finite then the value of a,b respectively is finite

Lt_(x to 0) (a sin x-sinx2x)/(tan^(3)x) is finite, then a=

lim_(x->0)(a sin x-2sin2x)/(tan^(3)x) is finite,then a=

If in a right angle DeltaABC,4 sin A cos B-1 =0 and tan A is finite, then