The value of tan left(frac{7 pi}{8}right) is | vedclass

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(B) Let $A = 	an \left(\frac{7 \pi}{8}ight)$.Since $\frac{7 \pi}{8} = \pi - \frac{\pi}{8}$,we have $A = 	an \left(\pi - \frac{\pi}{8}ight) = -

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$1-\sqrt{2}$

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(B) Let $A = 	an \left(\frac{7 \pi}{8}ight)$.Since $\frac{7 \pi}{8} = \pi - \frac{\pi}{8}$,we have $A = 	an \left(\pi - \frac{\pi}{8}ight) = -	an \left(\frac{\pi}{8}ight)$.Using the formula $	an(2	heta) = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,let $	heta = \frac{\pi}{8}$.Then $	an \left(\frac{\pi}{4}ight) = \frac{2 	an \left(\frac{\pi}{8}ight)}{1 - 	an^2 \left(\frac{\pi}{8}ight)} = 1$.Substituting $	an \left(\frac{\pi}{8}ight) = -A$,we get $\frac{2(-A)}{1 - (-A)^2} = 1$.$-2A = 1 - A^2 \Rightarrow A^2 - 2A - 1 = 0$.Using the quadratic formula $A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $A = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.Since $\frac{7 \pi}{8}$ lies in the $2^{nd}$ quadrant,$	an \left(\frac{7 \pi}{8}ight) Thus,$A = 1 - \sqrt{2}$.

The value of $\tan \left(\frac{7 \pi}{8}\right)$ is

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