Question

tan 3x tan 2x tan x=tan 3x -tan 2x -tan x

Answer 1

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

Consider,

$\rm :\longmapsto\:tan3x$

can be rewritten as

$\rm :\longmapsto\:tan3x = tan(2x + x)$

We know,

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

So, using this identity, we get

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

$\rm :\longmapsto\:tan3x - tan3xtan2xtanx = tan2x + tanx$

On transposition, we get

$\rm :\longmapsto\:tan3xtan2xtanx = tan3x - tan2x - tanx$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: tan3xtan2xtanx = tan3x - tan2x - tanx}}$

More to know :-

$\boxed{ \tt{ \: sin(x + y) = sinxcosy + sinycosx}}$

$\boxed{ \tt{ \: sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \tt{ \: cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \tt{ \: cos(x - y) = cosxcosy + sinxsiny}}$

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

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

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

Consider,

$\rm :\longmapsto\:tan3x$

can be rewritten as

$\rm :\longmapsto\:tan3x = tan(2x + x)$

We know,

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

So, using this identity, we get

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

$\rm :\longmapsto\:tan3x - tan3xtan2xtanx = tan2x + tanx$

On transposition, we get

$\rm :\longmapsto\:tan3xtan2xtanx = tan3x - tan2x - tanx$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: tan3xtan2xtanx = tan3x - tan2x - tanx}}$

More to know :-

$\boxed{ \tt{ \: sin(x + y) = sinxcosy + sinycosx}}$

$\boxed{ \tt{ \: sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \tt{ \: cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \tt{ \: cos(x - y) = cosxcosy + sinxsiny}}$

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

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$