Question

show tan3x tan2x tanx= tan 3x-tan2x-tanx ​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

$\rm \: 3x = 2x + x$

So,

$\rm\implies \:tan 3x =tan( 2x + x )$

We know,

$\boxed{ \rm{ \:tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: }}$

So, using this result, we get

$\rm \: tan3x = \dfrac{tan2x + tanx}{1 - tan2x \: tanx}$

$\rm \: tan3x(1 - tan2x \: tanx) = tan3x + tan2x$

$\rm \: tan3x - tan3x \: tan2x \: tanx = tan3x + tan2x$

On rearranging the terms, we get

$\rm \: tan3x - tan2x - tanx \: = \: tan3x \: tan2x \: tanx$

Hence,

$\rm\implies \:\boxed{ \rm{tan3x - tan2x - tanx =tan3x \: tan2x \: tanx \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(x - y) = sinx \: cosy \: - \: siny \: cosx}\\ \\ \bigstar \: \bf{sin(x + y) = sinx \: cosy \: + \: siny \: cosx}\\ \\ \bigstar \: \bf{cos(x + y) = cosx \: cosy \: - \: sinx \: siny}\\ \\ \bigstar \: \bf{cos(x - y) = cosx \: cosy \:+\: siny \: sinx}\\ \\ \bigstar \: \bf{tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany} }\\ \\ \bigstar \: \bf{tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

Solution :

$\begin{array}{l} \rm \tan (3 x)=\tan (2 x+x) \\ \\ \rm \therefore \tan (3 x)=\dfrac{\tan 2 x+\tan x}{1-\tan 2 x \tan x} \\ \\ \rm \therefore \tan 3 x[1-\tan 2 x \tan x]=\tan 2 x+\tan x \\ \\ \rm\therefore \tan 3 x-\tan 3 x \tan 2 x \tan x=\tan 2 x+\tan x \\ \\ \rm \therefore \tan 3 x-\tan 2 x-\tan x=\tan 3 x \tan 2 x \tan x \\ \\ \rm \therefore \tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x \end{array}$