The value of the expression tan left(frac{1} | vedclass

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(B) Let $ 	heta = \cos ^{-1} \left(\frac{2}{\sqrt{5}}ight) $. Then $ \cos 	heta = \frac{2}{\sqrt{5}} $. We need to find $ 	an \left(\frac{	heta}

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$ \sqrt{5}-2 $

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(B) Let $ 	heta = \cos ^{-1} \left(\frac{2}{\sqrt{5}}ight) $. Then $ \cos 	heta = \frac{2}{\sqrt{5}} $. We need to find $ 	an \left(\frac{	heta}{2}ight) $. Using the half-angle formula,$ 	an \left(\frac{	heta}{2}ight) = \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} $. Substituting the value of $ \cos 	heta $: $ 	an \left(\frac{	heta}{2}ight) = \sqrt{\frac{1-\frac{2}{\sqrt{5}}}{1+\frac{2}{\sqrt{5}}}} = \sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}} $. Rationalizing the denominator: $ \sqrt{\frac{(\sqrt{5}-2)(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2)}} = \sqrt{\frac{(\sqrt{5}-2)^2}{5-4}} = \sqrt{(\sqrt{5}-2)^2} = \sqrt{5}-2 $. Thus,the value is $ \sqrt{5}-2 $.

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

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