Question

CALCULATI:
$tg \frac{\pi}{12} + tg \frac{5\pi}{12}$
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Answer 1

Răspuns:

$tg\frac{\pi }{6}=\frac{2tg\frac{\pi }{12} }{1-tq^2\frac{\pi }{12} } =\frac{1}{\sqrt{3} }$

$2tan\frac{\pi }{12}=1-tan^2\frac{\pi }{12}$

$tan^2\frac{\pi }{12}+2\sqrt{3}tg\frac{\pi }{12}-1=0$

$tg\frac{\pi }{12}= \frac{-2\sqrt{3}+4 }{2} =2-\sqrt{3}$(Deoarece tg este positiva in primul cadran)

$tg\frac{\pi }{12}=2-\sqrt{3}$

$tg\frac{5\pi }{12}=tg(\frac{4\pi }{12}+\frac{\pi }{12})=tg(\frac{\pi }{3}+\frac{\pi }{12} )$

$=\frac{tg\frac{\pi }{3}+tg\frac{\pi }{12} }{1-tg\frac{\pi }{3}*tg\frac{\pi }{12} }$

$=\frac{\sqrt{3}+2-\sqrt{3} }{1-\sqrt{3}(2-\sqrt{3}) }$

$=\frac{2}{4-2\sqrt{3} }$

$=\frac{1}{2-\sqrt{3} } =\frac{2+\sqrt{3} }{4-3}=\frac{2+\sqrt{3} }{1}=2+\sqrt{3}$

$tg\frac{\pi }{12}+tg\frac{5\pi }{12} =2-\sqrt{3}+2+\sqrt{3}=4$

Answer 2

$\it tg\dfrac{\pi}{12}=tg(\dfrac{\pi}{4}-\dfrac{\pi}{6})=\dfrac{tg\dfrac{\pi}{4}-tg\dfrac{\pi}{6}}{1+tg\dfrac{\pi}{4}\cdot tg\dfrac{\pi}{6}}=\dfrac{1-\dfrac{\sqrt3}{3}}{1+\dfrac{\sqrt3}{3}}=\dfrac{^{3-\sqrt3)}3-\sqrt3}{\ \ 3+\sqrt3}=\\ \\ \\ =\dfrac{(3-\sqrt3)^2}{9-3}=\dfrac{9-6\sqrt3+3}{6}=\dfrac{12-6\sqrt3}{6}=\dfrac{6(2-\sqrt3)}{6}=2-\sqrt3$

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$\it tg\dfrac{5\pi}{12}=ctg(\dfrac{\pi}{2}-\dfrac{5\pi}{12})=ctg\dfrac{\pi}{12}=\dfrac{1}{tg\dfrac{\pi}{12}}=\dfrac{1}{2-\sqrt3}=2+\sqrt3$

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$\it tg\dfrac{\pi}{12}+tg\dfrac{5\pi}{12}=2-\sqrt3+2+\sqrt3=4$