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

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

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$2$

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(C) Let $	heta = 	an ^{-1}\left(\frac{\sqrt{5}-1}{2}ight)$. Then $	an 	heta = \frac{\sqrt{5}-1}{2}$.We need to find $	an(2	heta)$.Using the formula $	an(2	heta) = \frac{2 	an 	heta}{1 - 	an^2 	heta}$:$	an(2	heta) = \frac{2 \left(\frac{\sqrt{5}-1}{2}ight)}{1 - \left(\frac{\sqrt{5}-1}{2}ight)^2}$$= \frac{\sqrt{5}-1}{1 - \frac{5 + 1 - 2\sqrt{5}}{4}}$$= \frac{\sqrt{5}-1}{1 - \frac{6 - 2\sqrt{5}}{4}}$$= \frac{\sqrt{5}-1}{\frac{4 - 6 + 2\sqrt{5}}{4}}$$= \frac{4(\sqrt{5}-1)}{2\sqrt{5}-2}$$= \frac{4(\sqrt{5}-1)}{2(\sqrt{5}-1)}$$= 2$.

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

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