Question

Se tg x = 2/5, calcule tg (2x)

Answer 1

A tangente da soma de dois arcos (a+b) é dada por:

$\boxed{\sf \sf tg(a+b)~=~\dfrac{tg(a)~+~tg(b)}{1~-~tg(a)\cdot tg(b)}}$

Podemos reescrever tg(2x) como tg(x+x) e, portanto, temos que a tangente do arco 2x é dada por:

$\sf tg(2x)~=~tg(x+x)\\\\\\tg(2x)~=~\dfrac{tg(x)~+~tg(x)}{1-tg(x)\cdot tg(x)}\\\\\\Substituindo ~os ~valores:\\\\\\tg(2x)~=~\dfrac{\dfrac{2}{5}~+~\dfrac{2}{5}}{1~-~\dfrac{2}{5}\cdot \dfrac{2}{5}}\\\\\\tg(2x)~=~\dfrac{\dfrac{2+2}{5}}{1~-~\dfrac{2\cdot 2}{5\cdot 5}}\\\\\\tg(2x)~=~\dfrac{\dfrac{4}{5}}{1~-~\dfrac{4}{25}}\\\\\\tg(2x)~=~\dfrac{\dfrac{4}{5}}{\dfrac{25-4}{25}}\\\\\\tg(2x)~=~\dfrac{\dfrac{4}{5}}{\dfrac{21}{25}}\\\\\\tg(2x)~=~\dfrac{4}{5}\cdot \dfrac{25}{21}$

$\sf tg(2x)~=~\dfrac{4\cdot 25}{5\cdot 21}\\\\\\\sf tg(2x)~=~\dfrac{100}{105}\\\\\\\boxed{\sf tg(2x)~=~\dfrac{20}{21}}$

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