How do you solve $\frac{1 + tan θ}{1 - tan θ} = \frac{1 - tan θ}{1 + tan θ}$ $(1+tan theta)/(1-tan theta) = (1-tan theta)/(1+tan theta)$ ?

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Answer 1 · 2017-01-21T21:37:08

$θ = n π$ $theta = npi$ for any integer $n$ $n$

Let $t = tan θ$ $t = tan theta$

Then we want to solve:

$\frac{1 + t}{1 - t} = \frac{1 - t}{1 + t}$ $(1+t)/(1-t) = (1-t)/(1+t)$

Multiply both sides by $(1 - t) (1 + t)$ $(1-t)(1+t)$ to get:

${(1 + t)}^{2} = {(1 - t)}^{2}$ $(1+t)^2 = (1-t)^2$

Expand to get:

$1 + 2 t + t^{2} = 1 - 2 t + t^{2}$ $1+2t+t^2 = 1-2t+t^2$

Subtract $1 + t^{2}$ $1+t^2$ from both sides to get:

$2 t = - 2 t$ $2t = -2t$

Add $2 t$ $2t$ to both sides to get:

$4 t = 0$ $4t = 0$

Hence:

$t = 0$ $t = 0$

So it is necessary and sufficient that $tan θ = 0$ $tan theta = 0$

Note that $tan 0 = 0$ $tan 0 = 0$ and $tan$ $tan$ has period $π$ $pi$

So there are solutions::

$θ = n π$ $theta = npi$ for any integer $n$ $n$

How do you solve (1+tan theta)/(1-tan theta) = (1-tan theta)/(1+tan theta)1+tanθ1−tanθ=1−tanθ1+tanθ(1+tan theta)/(1-tan theta) = (1-tan theta)/(1+tan theta) ?