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 ?


Here's a nice (if I say so myself) geometric solution. Consider the right triangle $\triangle ABC$ with right $\angle ABC$, $|BC| = 3$ and $|AC| = 4$. $AD$ is the angle bisector of $\angle BAC$. Let $|BD| = x$ and hence $|CD| = 3-x$.
Now $\displaystyle |AB| = \sqrt{4^2 - 3^2} = \sqrt 7$ (Pythagoras). Note that $\displaystyle \theta = \frac 12 \arcsin \frac 34$ and $\tan \theta$ is the required ratio.
By the angle bisector theorem, $\displaystyle \frac x{3-x} = \frac{\sqrt 7}4$, giving $\displaystyle x = \frac{3\sqrt 7}{4 + \sqrt 7}$.
Hence the required ratio is $\displaystyle \tan \theta = \frac x{\sqrt 7} = \frac 3{4 + \sqrt 7} = \frac{4-\sqrt 7}3$ after rationalising the denominator.
