tan(2x) = 2tanx/(1 - tan2x) So tan(2x) +1 = 2tanx/(1 - tan2x) + 1 = 2tanx/(1 - tan2x) + (1 - tan2x)/(1 - tan2x) = (-tan2x + 2tanx + 1)/(1 - tan2x)
It depends on the domain of x.
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
I assume you mean (tanx+1)^2 In which case, (tanx+1)^2=tan2x+2tanx+1
The period of the tangent function is PI. The period of y= tan(2x) is PI over the coefficient of x = PI/2
tan(2x) = 2tanx/(1 - tan2x) So tan(2x) +1 = 2tanx/(1 - tan2x) + 1 = 2tanx/(1 - tan2x) + (1 - tan2x)/(1 - tan2x) = (-tan2x + 2tanx + 1)/(1 - tan2x)
there is non its an infinite line.
It depends on the domain of x.
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
By using the chain rule: dy/dx = dy/du x du/dx With y = tan2x Let u = tan x Then: y = u2 du/dx = d/dx tan x = sec2x dy/dx = dy/du x du/dx = 2u sec2x = 2 tan x sec2x
I assume you mean (tanx+1)^2 In which case, (tanx+1)^2=tan2x+2tanx+1
The period of the tangent function is PI. The period of y= tan(2x) is PI over the coefficient of x = PI/2
Oh, dude, I got you! So, like, a regular solid is just a three-dimensional shape where all the faces are congruent and all the angles are equal. It's like the cool kid on the block, you know? It's all uniform and symmetrical, just vibing in its own little geometric world.
well, they change in an organism and are influenced by differentiate
Differentiate Fatigue & Boredom?
differentiate articulation from enunciation?
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.