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**Hint:**Here, in the question, we have been given an equation and asked to prove that the left hand side of this equation is equal to the right hand side of the equation. While we will go through the solution we will get two values and one of them would be ignored, so we have to state the reason. We will start taking LHS and then simplify it using applicable identities and reach the RHS.

**Formulae used:**

The roots of the Quadratic equation \[a{x^2} + bx + c = 0\] is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

\[\sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}\]

**Complete step-by-step solution:**

We have to prove \[\tan \left( {\dfrac{1}{2}{{\sin }^{ - 1}}\dfrac{3}{4}} \right) = \dfrac{{4 - \sqrt 7 }}{3}\]

Firstly, Let’s assume \[\dfrac{1}{2}{\sin ^{ - 1}}\dfrac{3}{4} = \theta \]

\[ \Rightarrow {\sin ^{ - 1}}\dfrac{3}{4} = 2\theta \]

Taking \[\sin \] both sides, we get,

\[\sin \left( {{{\sin }^{ - 1}}\dfrac{3}{4}} \right) = \sin 2\theta \]

Using the property \[\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta \], we get,

\[\dfrac{3}{4} = \sin 2\theta \]

Now, using the identity\[\sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}\], we get,

\[\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \dfrac{3}{4}\]

Solving it further, we get

\[

8\tan \theta = 3 + 3{\tan ^2}\theta \\

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

Now, let’s assume \[\tan \theta = z\]

\[\therefore 3{z^2} - 8z + 3 = 0\]

Using the Discriminant method to solve for the value of \[z\], we get,

\[z = \dfrac{{ - \left( { - 8} \right) \pm \sqrt {{{\left( { - 8} \right)}^2} - 4\left( 3 \right)\left( 3 \right)} }}{{2\left( 3 \right)}}\]

Now, the simplest form of \[z\] can be calculated as

\[

z = \dfrac{{8 \pm \sqrt {28} }}{6} \\

\Rightarrow z = \dfrac{{4 \pm \sqrt 7 }}{3} \]

Now putting back the value \[\tan \theta = z\], we get

\[\tan \theta = \dfrac{{4 \pm \sqrt 7 }}{3}\]

Now, looking at the range of \[{\sin ^{ - 1}}x\], we obtain,

\[ - \dfrac{\pi }{2} \leqslant {\sin ^{ - 1}}\dfrac{3}{4} \leqslant \dfrac{\pi }{2}\]

Dividing this equation by 2, we get

\[ - \dfrac{\pi }{4} \leqslant \dfrac{1}{2}{\sin ^{ - 1}}\dfrac{3}{4} \leqslant \dfrac{\pi }{4}\]

Now, Putting the value of \[\dfrac{1}{2}{\sin ^{ - 1}}\dfrac{3}{4} = \theta \], we get

\[ - \dfrac{\pi }{4} \leqslant \theta \leqslant \dfrac{\pi }{4}\]

Taking \[\tan \]in this equation

\[\tan \left( { - \dfrac{\pi }{4}} \right) \leqslant \tan \left( \theta \right) \leqslant \tan \left( {\dfrac{\pi }{4}} \right) \\

\Rightarrow - 1 \leqslant \tan \theta \leqslant 1 \]

We can see that the value of \[\tan \theta \] must be less than or equal to \[1\] only.

But \[\dfrac{{4 + \sqrt 7 }}{3} > 1\], Hence, we will ignore this value

\[ \Rightarrow \tan \theta = \dfrac{{4 - \sqrt 7 }}{3}\]

Now, putting the values of \[\dfrac{1}{2}{\sin ^{ - 1}}\dfrac{3}{4} = \theta \], we get

\[\tan \left( {\dfrac{1}{2}{{\sin }^{ - 1}}\dfrac{3}{4}} \right) = \dfrac{{4 - \sqrt 7 }}{3}\]

**Hence, we have proved \[\tan \left( {\dfrac{1}{2}{{\sin }^{ - 1}}\dfrac{3}{4}} \right) = \dfrac{{4 - \sqrt 7 }}{3}\]**

**Note:**The symbol \[{\sin ^{ - 1}}\theta \] should not be confused with \[{\left( {\sin \theta } \right)^{ - 1}}\]. Infact \[{\sin ^{ - 1}}\theta \] is an angle, the value of whose sine is \[\theta \], similarly for other trigonometric functions.

The smallest numerical value of \[\theta \], either positive or negative, is known as the Principal Value of the function. The principal value of inverse of sine function must lie between \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\].

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