Question

simplificando a expressão tg X × cos X × sec X × cossec X obteremos:​

Answer 1

Sabendo que:

$\boxed{\sf tg(x)~=~\dfrac{sen(x)}{cos(x)}}\boxed{\sf sec(x)~=~\dfrac{1}{cos(x)}}\boxed{\sf cossec(x)~=~\dfrac{1}{sen(x)}}$

Temos:

$\sf tg(x)\cdot cos(x)\cdot sec(x)\cdot cossec(x)~=~\dfrac{sen(x)}{cos(x)}\cdot cos(x)\cdot \dfrac{1}{cos(x)}\cdot \dfrac{1}{sen(x)}\\\\\\\sf tg(x)\cdot cos(x)\cdot sec(x)\cdot cossec(x)~=~\dfrac{sen(x)\cdot cos(x)\cdot 1\cdot 1~\,}{cos(x)\cdot cos(x)\cdot sen(x)}$

$\sf tg(x)\cdot cos(x)\cdot sec(x)\cdot cossec(x)~=~\dfrac{\not\!\!\!sen(x)\cdot cos(x)}{cos(x)\cdot cos(x)\cdot \not\!\!\!sen(x)}\\\\\\\sf tg(x)\cdot cos(x)\cdot sec(x)\cdot cossec(x)~=~\dfrac{\not\!\!\! cos(x)}{cos(x)\cdot \not\!\!\!cos(x)}\\\\\\\sf tg(x)\cdot cos(x)\cdot sec(x)\cdot cossec(x)~=~\dfrac{1}{cos(x)}\\\\\\\boxed{\sf \sf tg(x)\cdot cos(x)\cdot sec(x)\cdot cossec(x)~=~sec(x)}$

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