The value of tan frac{pi}{8} is | vedclass

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(C) We know that $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$.Setting $	heta = \frac{\pi}{8}$,we have $	an \frac{\pi}{4} = \frac{2	an \fra

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$\sqrt{2}-1$

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(C) We know that $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$.Setting $	heta = \frac{\pi}{8}$,we have $	an \frac{\pi}{4} = \frac{2	an \frac{\pi}{8}}{1-	an^2 \frac{\pi}{8}}$.Since $	an \frac{\pi}{4} = 1$,let $y = 	an \frac{\pi}{8}$.Then $1 = \frac{2y}{1-y^2}$,which implies $1 - y^2 = 2y$,or $y^2 + 2y - 1 = 0$.Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$.Since $\frac{\pi}{8}$ is in the $1^{	ext{st}}$ quadrant,$	an \frac{\pi}{8} > 0$.Therefore,$	an \frac{\pi}{8} = \sqrt{2} - 1$.

The value of $\tan \frac{\pi}{8}$ is

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