tan left[ frac{1}{2} cos^{-1} left( frac{sqrt | vedclass

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Question

(A) Let $	heta = \frac{1}{2} \cos^{-1} \left( \frac{\sqrt{5}}{3} ight)$.Then $2	heta = \cos^{-1} \left( \frac{\sqrt{5}}{3} ight)$,which implies

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $	heta = \frac{1}{2} \cos^{-1} \left( \frac{\sqrt{5}}{3} ight)$.Then $2	heta = \cos^{-1} \left( \frac{\sqrt{5}}{3} ight)$,which implies $\cos 2	heta = \frac{\sqrt{5}}{3}$.Using the formula $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$,we have:$\frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} = \frac{\sqrt{5}}{3}$.Cross-multiplying gives $3 - 3 	an^2 	heta = \sqrt{5} + \sqrt{5} 	an^2 	heta$.Rearranging terms: $(3 + \sqrt{5}) 	an^2 	heta = 3 - \sqrt{5}$.$	an^2 	heta = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}$.Rationalizing the denominator: $	an^2 	heta = \frac{(3 - \sqrt{5})^2}{9 - 5} = \frac{(3 - \sqrt{5})^2}{4}$.Taking the square root: $	an 	heta = \frac{3 - \sqrt{5}}{2}$.

$\tan \left[ \frac{1}{2} \cos^{-1} \left( \frac{\sqrt{5}}{3} \right) \right] = $

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