# Writing $\tan^2(2\sec^{-1}(\frac{x}{3}))$ in algebraic form

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?

• Why do you need to use the double-angle formula? Just square the value of $\tan\theta$ that you’ve come up with. – amd Sep 30 '18 at 0:07
• Because there is a 2 inside the tan – Villa Sep 30 '18 at 0:09
• This is alright, but note that you have to plug $\tan{\theta}=\frac{\sqrt{x^2-9}}{3}$ – 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 $$\tan^2 (2\theta) = (\frac {2\tan \theta}{1 - \tan^2 \theta})^2 =$$
$$\frac {4\frac{x^2 - 9}9}{(1- \frac {x^2 -9}9)^2}=$$