Find the value of tan frac{pi}{8}. | vedclass

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(B) Let $x = \frac{\pi}{8}$. Then $2x = \frac{\pi}{4}$.Using the formula $	an 2x = \frac{2 	an x}{1 - 	an^2 x}$,we have:$	an \frac{\pi}{4} = \frac

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$\sqrt{2} - 1$

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(B) Let $x = \frac{\pi}{8}$. Then $2x = \frac{\pi}{4}$.Using the formula $	an 2x = \frac{2 	an x}{1 - 	an^2 x}$,we have:$	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}$$1 - y^2 = 2y$$y^2 + 2y - 1 = 0$Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$y = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$Since $\frac{\pi}{8}$ is in the first quadrant,$	an \frac{\pi}{8} > 0$. Therefore,$y = \sqrt{2} - 1$.

Find the value of $\tan \frac{\pi}{8}$.

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