Solve for x : tan(π + x) + tan(2π + x) + tan(3π + x) + ⋯ + tan(10

1.	Solve for x : tan(π + x) + tan(2π + x) + tan(3π + x) + ⋯ + tan(10π + x) = 10 , where, −2π ≤ x ≤ 2π
Answer» $\large\underline{\sf{Given- }}$ $\rm :\longmapsto\: - 2\pi \leqslant <klux>X</klux> \leqslant 2\pi$ and $\rm :\longmapsto\:tan(\pi + x) + tan(2\pi + x) + tan(3\pi + x) + - - - + tan(10\pi + x) = 10$ $\large\underline{\sf{To\:Find - }}$ The value of x. $\large\underline{\sf{Solution-}}$ Given that, $\rm :\longmapsto\:tan(\pi + x) + tan(2\pi + x) + tan(3\pi + x) + - - - + tan(10\pi + x) = 10$ We know that, $\boxed{ \rm{ tan(n\pi + x) = tanx \: \forall} \: n \: \in \: Z {}^{ + } }$ So, $\rm :\longmapsto\:tan(\pi + x) = tanx$ $\rm :\longmapsto\:tan(2\pi + x) = tanx$ $\rm :\longmapsto\:tan(3\pi + x) = tanx$ . . . . $\rm :\longmapsto\:tan(10\pi + x) = tanx$ So, given EQUATION $\rm :\longmapsto\:tan(\pi + x) + tan(2\pi + x) + tan(3\pi + x) + - - - + tan(10\pi + x) = 10$ can be REWRITTEN as $\rm :\longmapsto\:tanx + tanx + tanx + - - - - + tanx = 10$ $\rm :\longmapsto\:10tanx = 10$ $\rm :\longmapsto\:tanx = 1$ $\rm :\longmapsto\:tanx = tan\dfrac{\pi}{4}$ We know, $\boxed{ \rm{ tanx = tany \: \implies \: x = n\pi + y \: \forall \: n \: \in \: Z}}$ So, using this identity, we get $\rm :\implies\:x = n\pi + \dfrac{\pi}{4} \: \forall \: n \: \in \: Z$ As, it is given that, $\rm :\longmapsto\: - 2\pi \leqslant x \leqslant 2\pi$ So, x can take the values, $\bf\implies \:x = \dfrac{\pi}{4}, \: \dfrac{5\pi}{4}, \: - \dfrac{3\pi}{4}, \: - \dfrac{7\pi}{4}$ ADDITIONAL INFORMATION :- $\begin{gathered}\begin{gathered}\boxed{\begin{array}{c\|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \: \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \: \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \: \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \: \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \: \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \: \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$

Solve for x : tan(π + x) + tan(2π + x) + tan(3π + x) + ⋯ + tan(10π + x) = 10 , where, −2π ≤ x ≤ 2π

Answer»

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

$\rm :\longmapsto\: - 2\pi \leqslant <klux>X</klux> \leqslant 2\pi$

and

$\rm :\longmapsto\:tan(\pi + x) + tan(2\pi + x) + tan(3\pi + x) + - - - + tan(10\pi + x) = 10$

$\large\underline{\sf{To\:Find - }}$

The value of x.

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

Given that,

$\rm :\longmapsto\:tan(\pi + x) + tan(2\pi + x) + tan(3\pi + x) + - - - + tan(10\pi + x) = 10$

We know that,

$\boxed{ \rm{ tan(n\pi + x) = tanx \: \forall} \: n \: \in \: Z {}^{ + } }$

So,

$\rm :\longmapsto\:tan(\pi + x) = tanx$

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

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

$\rm :\longmapsto\:tan(10\pi + x) = tanx$

So, given EQUATION

$\rm :\longmapsto\:tan(\pi + x) + tan(2\pi + x) + tan(3\pi + x) + - - - + tan(10\pi + x) = 10$

can be REWRITTEN as

$\rm :\longmapsto\:tanx + tanx + tanx + - - - - + tanx = 10$

$\rm :\longmapsto\:10tanx = 10$

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

$\rm :\longmapsto\:tanx = tan\dfrac{\pi}{4}$

We know,

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

So, using this identity, we get

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

As, it is given that,

$\rm :\longmapsto\: - 2\pi \leqslant x \leqslant 2\pi$

So,

x can take the values,

$\bf\implies \:x = \dfrac{\pi}{4}, \: \dfrac{5\pi}{4}, \: - \dfrac{3\pi}{4}, \: - \dfrac{7\pi}{4}$

ADDITIONAL INFORMATION :-

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