Question

Simplificando $tg(\dfrac{\pi }{20} ).tg(\dfrac{3\pi }{20} ).tg(\dfrac{5\pi }{20} ).tg(\dfrac{7\pi }{20} ).tg(\dfrac{9\pi }{20} )$, obtém-se:


a) 1

b) -1

c) $\dfrac{\sqrt{3} }{3}$

d) $\sqrt{3}$

e) 0

Answer 1

Resposta:

Explicação passo-a-passo:

Lembrando que

$tg(\frac{\pi}{2}-\alpha)=cotg(\alpha)\\\\e\\cotg(\alpha)=\dfrac{1}{tg(\alpha)}=>cotg(\alpha).tg(\alpha)=1$

Podemos escrever

$tg(\dfrac{\pi}{20})=tg(\dfrac{10\pi-9\pi}{20})=tg(\dfrac{10\pi}{20}-\dfrac{9\pi}{20})=tg(\dfrac{\pi}{2}-\dfrac{9\pi}{20})=cotg(\dfrac{9\pi}{20})\\\\\\tg(\dfrac{3\pi}{20})=tg(\dfrac{10\pi-7\pi}{20})=tg(\dfrac{10\pi}{20}-\dfrac{7\pi}{20})=tg(\dfrac{\pi}{2}-\dfrac{7\pi}{20})=cotg(\dfrac{7\pi}{20})\\\\\\tg(\dfrac{5\pi}{20})=tg(\dfrac{\pi}{4})=1$

substituindo na expressão inicial

$cotg(\dfrac{9\pi}{20}).cotg(\dfrac{7\pi}{20}).1.tg(\dfrac{7\pi}{20}).tg(\dfrac{9\pi}{20})=\\\\\\rearrumando\\\\=cotg(\dfrac{9\pi}{20}).tg(\dfrac{9\pi}{20}).cotg(\dfrac{7\pi}{20}).tg(\dfrac{7\pi}{20})=1$