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Miscellaneous Exercise 3 (II) · Q115

Q.Prove the following: tan⁡π8=2−1\tan\dfrac{\pi}{8}=\sqrt2-1

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Step 1: Let y=tan⁡π8y=\tan\dfrac{\pi}{8}. Since 2×π8=π42\times\dfrac{\pi}{8}=\dfrac{\pi}{4}, use tan⁡2x=2tan⁡x1−tan⁡2x\tan2x=\dfrac{2\tan x}{1-\tan^2x}: 1=2y1−y21=\dfrac{2y}{1-y^2}.

Step 2: Cross-multiply: 1−y2=2y1-y^2=2y, i.e. y2+2y−1=0y^2+2y-1=0. …

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