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

Given that \( \tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right) \)
Let \( \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\theta \) then, \( \cos \theta=\frac{2}{\sqrt{5}} \)
So, \( \tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right)=\tan \left(\frac{1}{2} \theta\right) \)
We know that, \( \tan \theta=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}} \)
So, \( \tan \left(\frac{1}{2} \theta\right)=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \)
\( =\frac{\sqrt{1-\frac{2}{\sqrt{5}}}}{\sqrt{1+\frac{2}{\sqrt{5}}}}=\frac{\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+2}} \) \( =\frac{\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+2}} \times \frac{\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}-2}} \) \( =\frac{\sqrt{5}-2}{1}=\sqrt{5}-2 \)