ANSWER
[tex]\tan(\sin^{ - 1}( \frac{x}{2} )) = \frac{x}{ \sqrt{4 - {x}^{2} } } \: \:where \: \: x \ne \pm2[/tex]
EXPLANATION
We want to find the exact value of
[tex] \tan( \sin^{ - 1}( \frac{x}{2} ) ) [/tex]
Let
[tex]y = \sin^{ - 1}( \frac{x}{2} )[/tex]
This implies that
[tex] \sin(y) = \frac{x}{2} [/tex]
This implies that,
The opposite is x units and the hypotenuse is 2 units.
The adjacent side is found using Pythagoras Theorem.
[tex] {a}^{2} + {x}^{2} = {2}^{2} [/tex]
[tex]{a}^{2} + {x}^{2} = 4[/tex]
[tex]{a}^{2} = 4 - {x}^{2}[/tex]
[tex]a= \sqrt{4 - {x}^{2}} [/tex]
This implies that,
[tex] \tan(y) = \frac{opposite}{adjacent} [/tex]
[tex]\tan(y) = \frac{x}{ \sqrt{4 - {x}^{2} } } [/tex]
But
[tex]y = \sin^{ - 1}( \frac{x}{2} )[/tex]
This implies that,
[tex]\tan(\sin^{ - 1}( \frac{x}{2} )) = \frac{x}{ \sqrt{4 - {x}^{2} } } \: \:where \: \: x \ne \pm2[/tex]
