Inverse Trigonometric Functions

Evaluate: $\;$ $\tan^2 \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{2}{3}\right)\right]$

Let $\sin^{-1} \left(\dfrac{2}{3}\right) = \theta \;\;\; \text{ where } \theta \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\sin \left(\theta\right) = \dfrac{2}{3}$

$\therefore \; \tan^2 \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{2}{3}\right)\right] = \tan^2 \left(\dfrac{\theta}{2}\right)$

Now, $\;$ $\sin \left(\theta\right) = \dfrac{2 \tan \left(\dfrac{\theta}{2}\right)}{1 + \tan^2 \left(\dfrac{\theta}{2}\right)}$

i.e. $\;$ $\dfrac{2 \tan \left(\dfrac{\theta}{2}\right)}{1 + \tan^2 \left(\dfrac{\theta}{2}\right)} = \dfrac{2}{3}$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) - 3 \tan \left(\dfrac{\theta}{2}\right) + 1 = 0$

i.e. $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 \pm \sqrt{9 - 4}}{2}$

i.e. $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 + \sqrt{5}}{2}$ $\;$ OR $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 - \sqrt{5}}{2}$

$\therefore \;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \left(\dfrac{3 + \sqrt{5}}{2}\right)^2$ $\;$ OR $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \left(\dfrac{3 - \sqrt{5}}{2}\right)^2$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{14 + 6 \sqrt{5}}{4}$ $\;$ OR $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{14 - 6 \sqrt{5}}{4}$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{7 \pm 3 \sqrt{5}}{2}$

$\because \;$ $\dfrac{7 + 3 \sqrt{5}}{2} \; \notin \; \left[-1, 1\right]$ $\;$ and $\;$ $\dfrac{7 - 3 \sqrt{5}}{2} \; \in \; \left[-1, 1\right]$,

$\therefore \;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{7 - 3 \sqrt{5}}{2}$

i.e. $\;$ $\tan^2 \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{2}{3}\right)\right] = \dfrac{7 - 3 \sqrt{5}}{2}$