Question

Solve :- tanx + tan3x+ tanxtan3x=1

Please explain with steps. ​

Answer 1

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

Given Trigonometric equation is

$\rm :\longmapsto\:tanx + tan3x + tanx \: tan3x = 1$

can be rewritten as

$\rm :\longmapsto\:tanx + tan3x = 1 - tanx \: tan3x$

can be further rewritten as

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

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:tan(x + 3x) = 1$

$\rm :\longmapsto\:tan(4x) = 1$

$\rm :\longmapsto\:tan4x= tan\bigg[\dfrac{\pi}{4} \bigg]$

We know,

$\boxed{ \tt{ \: tanx = tany \bf\implies \:\sf x = n\pi + y \: \forall \: n \in \: Z}}$

So, using this identity, we get

$\bf\implies \: 4x = n\pi + \dfrac{\pi}{4} \: \forall \: n \in \: Z$

$\red{\bf\implies \: \boxed{ \tt{ \: x = \dfrac{n\pi}{4} + \dfrac{\pi}{16} \: \forall \: n \in \: Z}}}$

More to know :-

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