#10; tan(13π/12) =tan(π+(π/12)) = tan(π/12) = tan((π/4) (π/6)) = [(tan(π/4) tan(π/6)/1+tan(π/4)>tan(π/6))] [(1 (1/√(3))/1+1×(1/√( ...

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Introduction

tan 13 π/ 12 ...

Question

$tan \frac{13 π}{12} = ?$

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Solution

$t a n (\frac{13 π}{12}) = t a n (π + (\frac{π}{12})) = - t a n (\frac{π}{12}) = - t a n ((\frac{π}{4}) - (\frac{π}{6})) = - [(\frac{t a n (\frac{π}{4}) - t a n (\frac{π}{6})}{1 + t a n (\frac{π}{4}) t a n (\frac{π}{6})})] - [(\frac{1 - (\frac{1}{\sqrt{3}})}{1 + 1 \times (\frac{1}{\sqrt{3}})})] = - ⎡ ⎢ ⎣ (\frac{(\frac{\sqrt{3} - 1}{\sqrt{3}})}{(\frac{\sqrt{3} + 1}{\sqrt{3}})}) ⎤ ⎥ ⎦ = - (\frac{(\sqrt{3} - 1)}{\sqrt{3} + 1})$

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