Show that tan3x. tan2x. tanx = tan3x - tan2x -tanx
Answers
EXPLANATION.
⇒ tan3x.tan2x.tanx = tan3x - tan2x - tan x.
As we know that,
⇒ tan3x = tan2x + tan x.
⇒ tan3x = tan(2x + x).
As we know that,
Formula of :
⇒ tan(x + y) = (tan x + tan y)/(1 - tan x. tan y).
Using this formula in the equation, we get.
⇒ tan3x = (tan2x + tan x)/(1 - tan2x.tanx).
⇒ tan3x[1 - tan2x.tanx] = tan2x + tan x.
⇒ tan3x - tan3x.tan2x .tan x = tan2x + tan x.
⇒ tan3x - tan3x.tan2x.tan x - tan2x - tan x = 0.
⇒ tan3x - tan2x - tan x = tan3x.tan2x.tan x.
Hence Proved.
MORE INFORMATION.
Trigonometrical ratios of multiple angles.
(1) = sin2θ = 2sinθ.cosθ = 2tanθ/1 + tan²θ.
(2) = cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ = 1 - tan²θ/1 + tan²θ.
(3) = tan2θ = 2tanθ/1 - tan²θ.
(4) = sin3θ = 3sinθ - 4sin³θ.
(5) = cos3θ = 4cos³θ - 3cosθ.
(6) = tan3θ = 3tanθ - tan³θ/1 - 3tan²θ.
Step-by-step explanation:
tan3x=tan(2x+x)
or, tan3x= tan 2x+tan x/1-tan 2.tan x
or, tan3x−tan3x.tan2x.tanx=tan2x+tanx
or, tan3x−tan2x−tanx=tan3x.tan2x.tanx.