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I have $$\tan (2\alpha) = \frac {4n^2}{4n^4-1}$$ And I want to solve for $\alpha$. So far I have tried applying the inverse tangent to both sides and dividing by two, but the book says that the answer should be $\frac {1}{2n^2}$ .
So then I use the double angle formulas for sine and cosine and after some rearraging I get to this:$$2n^2-\frac {1}{2n^2}=\cot(\alpha) -\tan(\alpha)$$ Which seems to imply what the book says but I am unable to get rid of the cotangent to get a rigorous result. How should I attack the problem?

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$$ \underbrace{\frac {2\tan\alpha}{1-\tan^2\alpha} = \tan(2\alpha)}_\text{double-angle formula} = \underbrace{\frac{4n^2}{4n^4-1} = \frac{2\left( \dfrac 1 {2n^2} \right)}{1 - \left(\dfrac 1{2n^2}\right)^2}}_{\begin{smallmatrix} \text{This gets us a “1'' where we} \\ \text{need it in the denominator.} \end{smallmatrix}}. $$

We have $\dfrac{2\tan\alpha}{1-\tan^2\alpha} = \dfrac{2\left( \dfrac 1 {2n^2} \right)}{1 - \left(\dfrac 1{2n^2}\right)^2}\quad$ if $\quad\tan\alpha = \dfrac 1{2n^2}$.

PS in response to comments: Two issues arise.

First, if $\dfrac{2x}{1-x^2} = \dfrac{2y}{1-y^2}$, does it follow that $x=y$? If not, we could have $\tan\alpha=x$ and $\dfrac 1 {2n^2} = y$; thus where I said $\text{“}$If $x=y\text{''}$, one could not then add $\text{“}$only if$\text{''}$. Then one would have to check for extraneous roots. One way to deal with this is to actually check for extraneous roots.

In fact the equation $\dfrac{2x}{1-x^2} = \dfrac{2y}{1-y^2}$ is equivalent to

$$ 2x(1-y^2) = 2y(1-x^2) \text{ and } x,y\notin\{-1,1\}, $$ unless you want to allow $\infty$ as a value (which might not be a bad idea) in which case you can drop the part including and after the word $\text{“and''}$. This is a quadratic equation and if you solve it for $x$ in terms of $y$ (or vice-versa) you do get two distinct solutions. Howeover, both of the resulting distinct values of $\alpha$ lead to the same value of $2\alpha$, and therefore to a solution of the original equation.

Second: Do we only want solutions in the interval $(-\pi/2,\pi/2)$? If so, then you don't have to list infinitely many solutions; otherwise you do.

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    $\begingroup$ Clever; I like it! Notice $\frac{2x}{1-x^2}$ is one-to-one for $x>0$. $\endgroup$ – Fimpellizieri May 20 '16 at 0:40
  • $\begingroup$ I like it, but when can I use this method? The trouble I was having earlier was that just because the terms in an equation seem to be playing the same roles does not mean they are equal e.g. $4+3=5+2$ but that does not mean $5=4$. I just want to know why that is the unique solution. $\endgroup$ – Guacho Perez May 20 '16 at 4:48
  • $\begingroup$ @GuachoPerez : I've added a postscript to the question to address this. $\qquad$ $\endgroup$ – Michael Hardy May 20 '16 at 21:17
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You have $$\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{4n^2}{4n^4-1}$$ Rearranging this gives $$\tan^2\alpha+\tan\alpha\left(2n^2-\frac{1}{2n^2}\right)-1=0$$ $$\Rightarrow\left(\tan\alpha+2n^2\right)\left(\tan\alpha-\frac{1}{2n^2}\right)=0$$

Hence $$\tan\alpha=-2n^2 \text{or} \frac{1}{2n^2}$$

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