Stumped here, I've tried and tried. Here's where I am

Put this expression into algebraic form: $$\tan^2\left(2\sec^{-1}\left(\frac{x}{3}\right)\right)$$

Let $\theta=\sec^{-1}(\frac{x}{3})$ thus $\sec \theta=\frac{x}{3}$ $cos \theta=\frac{3}{x}$ since $\sec = 1/\cos$.

This angle must exist within the 1st and 2nd quadrants because thats where arcsec is defined right?

So then I get opposite side = $3$, hypotenuse = $x$, adjacent = $\sqrt{x^2-3}$. And I know

$$\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$$

Am I on the right track? Now do I plug in $\tan\theta=\frac{\sqrt{x^2-3}}{3}$ and solve or have I missed something?

  • $\begingroup$ Why do you need to use the double-angle formula? Just square the value of $\tan\theta$ that you’ve come up with. $\endgroup$ – amd Sep 30 '18 at 0:07
  • $\begingroup$ Because there is a 2 inside the tan $\endgroup$ – Villa Sep 30 '18 at 0:09
  • $\begingroup$ This is alright, but note that you have to plug $\tan{\theta}=\frac{\sqrt{x^2-9}}{3}$ $\endgroup$ – Villa Sep 30 '18 at 0:18

You have $\cos \theta = \frac 3x$. So $\sin \theta = \sqrt{1 - \frac 9{x^2}}$ and $\tan \theta = \frac {\sqrt{1 - \frac 9{x^2}}}{\frac 3x} = \frac {\sqrt{x^2 -9}}{3}$

So $\tan^2 (2\theta) = (\frac {2\tan \theta}{1 - \tan^2 \theta})^2 =$

$\frac {4\frac{x^2 - 9}9}{(1- \frac {x^2 -9}9)^2}=$


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