The value of tan left( frac{1}{2} cos^{-1} le | vedclass

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Question

(C) Let $\cos^{-1} \left( \frac{\sqrt{5}}{3} \right) = \alpha$. Then $\cos \alpha = \frac{\sqrt{5}}{3}$,where $0 < \alpha < \frac{\pi}{2}$. We need to

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$\frac{3 - \sqrt{5}}{2}$

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(C) Let $\cos^{-1} \left( \frac{\sqrt{5}}{3} ight) = \alpha$. Then $\cos \alpha = \frac{\sqrt{5}}{3}$,where $0 We need to find $	an \left( \frac{\alpha}{2} ight)$. Using the half-angle formula,$	an \left( \frac{\alpha}{2} ight) = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}$. Substituting the value of $\cos \alpha$: $	an \left( \frac{\alpha}{2} ight) = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{1 + \frac{\sqrt{5}}{3}}} = \sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}}$. To rationalize the denominator,multiply the numerator and denominator by $(3 - \sqrt{5})$: $\sqrt{\frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}} = \sqrt{\frac{(3 - \sqrt{5})^2}{9 - 5}} = \sqrt{\frac{(3 - \sqrt{5})^2}{4}} = \frac{3 - \sqrt{5}}{2}$.

The value of $\tan \left( \frac{1}{2} \cos^{-1} \left( \frac{\sqrt{5}}{3} \right) \right)$ is

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