<s>[INST] How to prove $\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}$ Prove

$$
\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}
$$
  and justify why $\frac{4+\sqrt{7}}{3}$ is ignored.

My Attempt:
$$
\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}\implies\tan x-\tan x\tan^2\frac{x}{2}=2\tan\frac{x}{2}\\
\implies \tan^2\frac{x}{2}(\tan x)+2\tan\frac{x}{2}-\tan x=0\\
\implies \tan\frac{x}{2}=\frac{-2\pm2\sec x}{2\tan x}=\frac{-1\pm\sec x}{\tan x}
$$
Using this,
$$
\tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-1\pm\frac{4}{\sqrt{7}}}{\frac{3}{\sqrt{7}}}=\frac{-\sqrt{7}\pm4}{3}
$$
As $0\leq\frac{\sin^{-1}3/4}{2}\leq\frac{\pi}{4}\implies0\leq\tan(\frac{\sin^{-1}3/4}{2})\leq1$
$$
\implies \tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-\sqrt{7}+4}{3}
$$
Is my attempt correct and where is the value $\frac{4+\sqrt{7}}{3}$ to be excluded ?
 [/INST] The method you used to eliminate $\frac{-4-\sqrt7}{3}$ is correct. Here I want to explain why it comes out to be a root of the equation so that things become more clear.
Consider the following 
$$\tan(n\pi +2\theta)=\tan2\theta$$
$$\frac{\tan(\frac{n\pi}{2}+\theta)}{1-\tan^2(\frac{n\pi}{2}+\theta)}=\tan2\theta$$
Now it follows that not only $\tan\theta$ but also $\tan(\frac{n\pi}{2}+\theta)$ is a root of $$\tan2\theta=\frac{x}{1-x^2}$$ and to decide which of the two roots of the quadratic equation is of $\tan\theta$ is to be determined by the value of $2\theta$ and not by merely knowing the value of $\tan2\theta$
</s>