What is the value of $tan\bigg(\frac{1}{2}sin^{-1}\frac{3}{4}\bigg)$

Toolbox:
• $sin^{-1}x = \tan^{-1}\large \frac{x}{\sqrt (1-x^2)}$
• $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$
Given $tan\bigg(\frac{1}{2}sin^{-1}\frac{3}{4}\bigg)$
$\textbf{Step 1}$:
Let $\theta = sin^{-1}\frac{3}{4} \rightarrow tan (\frac{1}{2}sin^{-1}\frac{3}{4}) = tan \frac{\theta}{2}$
We know that $sin^{-1}x = \tan^{-1}\large \frac{x}{\sqrt (1-x^2)}$
Therefore, if we assume that $\theta = sin^{-1}\frac{3}{4}$ this is $= \tan^{-1}\large \frac{\large \frac{3}{4}}{\sqrt (1-(\large \frac{3}{4})^2)}$
$\Rightarrow \theta $$= \tan^{-1}\large \frac{\large \frac{3}{4}}{\sqrt \large\frac{16-9}{16}}$$ = tan^{-1} (\frac{3}{4} \times \frac{\sqrt16}{\sqrt7})$
$\Rightarrow \theta = tan^{-1} \frac{3}{\sqrt7}$
$\Rightarrow tan \theta = \frac{3}{\sqrt7}$
$\textbf{Step 2}$:
We know that $\tan x=\large \frac{2\tan\frac{x}{2}}{1-tan^2\frac{x}{2}}$
$\Rightarrow tan \theta =\large \frac{2\tan\frac{\theta}{2}}{1-tan^2\frac{\theta}{2}}$$= \frac{3}{\sqrt 7}$
$\Rightarrow 3-3\tan^2\frac{\theta}{2}=\frac{2}{\sqrt 7} \tan\frac{\theta}{2}$
$\Rightarrow$ $3\tan^2\frac{\theta}{2}+2\sqrt 7\tan\frac{\theta}{2}-3=0$
$\Rightarrow$ $\tan\frac{\theta}{2}=\large \frac{-2\sqrt 7\pm\sqrt(28+36)}{6}$
$\Rightarrow$ $\tan\frac{\theta}{2}=\large \frac{-2\sqrt 7\pm 8}{6} = \large \frac{-\sqrt 7\pm 4}{3}$
Since $sin\frac{\theta}{2}$ is acute, $tan\frac{\theta}{2}=\frac{4-\sqrt 7}{3}$