Question

(CEPBJ) Sabendo que cotgx = 1/2, o valor da tg 2x é igual a

Answer 1

Condição I:

$cotg\,x=\frac{1}{2}\\\\\frac{\cos\,x}{\sin\,x}=\frac{1}{2}\\\\\frac{\cos\,x}{\sin\,x}=\frac{k}{2k}\Rightarrow\begin{cases}\cos\,x=k\\\sin\,x=2k\end{cases}$
 
 Por fim,

$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}\\\\\tan(2x)=\frac{2\cdot\sin\,x\cdot\cos\,x}{\cos^2x-\sin^2x}\\\\\tan(2x)=\frac{2\cdot2k\cdot\,k}{k^2-4k^2}\\\\\tan(2x)=\frac{4k^2}{-3k^2}\\\\\boxed{\tan(2x)=-\frac{4}{3}}$

Answer 2

$cotg~x=1/2\\tg~x=2/1\\tg~x=2$

$tg~(a+b)=(tg~a+tg~b)/(1-tg~a*tg~b)\\tg~(x+x)=(tg~x+tg~x)/(1-tg~x*tg~x)\\\\\boxed{tg~2x=\dfrac{2*tg~x}{1-tg^{2}x}}$
________

$tg~2x=\dfrac{2*2}{1-2^{2}}~~\therefore~~\dfrac{4}{1-4}~~\therefore~~\dfrac{4}{-3}\\\\\\\boxed{\boxed{tg~2x=-\frac{4}{3}}}$