Show that $\tan \left( {\frac{1}{2}{{\sin }^{ - 1}}\frac{3}{4}} \right) = \frac{{4 - \sqrt 7 }}{3}$ and justify why the other value $\frac{{4 + \sqrt 7 }}{3}$ is ignored?
[NCERT,Exemplar.2.3,Q.18,Page.37]

We have, $\tan \left( {\frac{1}{2}{{\sin }^{ - 1}}\frac{3}{4}} \right) = \frac{{4 - \sqrt 7 }}{3}$

$therefore, $

${\rm{LHS}} = \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{3}{4}} \right)} \right]$

Let $\frac{1}{2}{\sin ^{ - 1}}\frac{3}{4} = \theta \Rightarrow {\sin ^{ - 1}}\frac{3}{4} = 2\theta $

$ \Rightarrow $$\sin 2\theta = \frac{3}{4} \Rightarrow \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \frac{3}{4}$

$ \Rightarrow $$3 + 3{\tan ^2}\theta = 8\tan \theta $
$ \Rightarrow $$3{\tan ^2}\theta - 8\tan \theta + 3 = 0$

Let $\tan \theta = y$

Therefore equation will become : $3{y^2} - 8y + 3 = 0$

$ \Rightarrow $$y = \frac{{ + 8 \pm \sqrt {64 - 4 \times 3 \times 3} }}{{2 \times 3}} = \frac{{8 \pm \sqrt {28} }}{6}$
$ = \frac{{2[4 \pm \sqrt 7 ]}}{{2 \cdot 3}}$
$ \Rightarrow $$\tan \theta = \frac{{4 \pm \sqrt 7 }}{3}$

$ \Rightarrow $$\theta = {\tan ^{ - 1}}\left[ {\frac{{4 \pm \sqrt 7 }}{3}} \right]$
$\left\{ {{\rm{but}}\frac{{4 + \sqrt 7 }}{3} > \frac{1}{2} \cdot \frac{\pi }{2}} \right.$,
since $\left. {\max \left[ {\tan \left( {\frac{1}{2}{{\sin }^{ - 1}}\frac{3}{4}} \right)} \right] = 1} \right\}$

$therefore, $${\rm{LHS}} = \tan {\tan ^{ - 1}}\left( {\frac{{4 - \sqrt 7 }}{3}} \right) = \frac{{4 - \sqrt 7 }}{3} = {\rm{RHS}}$

Note Since, $ - \frac{\pi }{2} \le {\sin ^{ - 1}}\frac{3}{4} \le \pi /2$$ \Rightarrow $$\frac{{ - \pi }}{4} \le \frac{1}{2}{\sin ^{ - 1}}\frac{3}{4} \le \pi /4$

$therefore, $$\tan \left( {\frac{{ - \pi }}{4}} \right) \le \tan \frac{1}{2}\left( {{{\sin }^{ - 1}}\frac{3}{4}} \right) \le \tan \frac{\pi }{4}$

$ \Rightarrow $$ - 1 \le \tan \left( {\frac{1}{2}{{\sin }^{ - 1}}\frac{3}{4}} \right) \le 1$

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