Question

Sabendo -se que ( imagem) o valor da expressão E = M2 - N2 é igual a:

Answer 1

$\displaystyle \text M = \frac{[\text{tg}^260^\circ]^{\text n}+[\text{cotg}^230^\circ]^{-\text n}}{\text{sec }60^\circ} \ ; \ \text N = \frac{[\text{cotg}^230^\circ]^{\text n}-[\text{tg}^260^\circ]^{-\text n}}{\text{cossec }30^\circ} \\\\\\ \text E =\text M^2-\text N^2=\ ?$

Primeiro vamos colocar os valores numéricos para agilizar nossa vida :

$\text {tg}^260^\circ = \text{cotg}^230^\circ = (\sqrt{3})^2 = 3 \\\\ \text{sec }60^\circ=\text{cossec }60^\circ =2$

Daí  :

$\displaystyle \text M = \frac{3^{\text n}+3^{-\text n}}{2} \ ; \ \text N = \frac{3^{\text n}-3^{-\text n}}{2} \\\\\\ \text M = \frac{1}{2}(3^{\text n}+3^{-\text n}) \to \text M = \frac{1}{2}(3^{\text n}+\frac{1}{3^{\text n}}) \to \text M = \frac{(3^{\text n})^2+1}{2.3^{\text n} } \\\\\\ \text N = \frac{1}{2}(3^{\text n}-3^{-\text n}) \to \text N = \frac{1}{2}(3^{\text n}-\frac{1}{3^{\text n}}) \to \text N = \frac{(3^{\text n})^2-1}{2.3^{\text n} } \\\\\\ \underline{\text{temos}} :$

$\displaystyle \text E = (\frac{(3^{\text n})^2+1}{2.3^{\text n} })^2-(\frac{(3^{\text n})^2-1}{2.3^{\text n} })^2 \\\\\\ \underline{\text{Fa{\c c}amos}}: 3^{\text n} = \text x \\\\\\ \text E = (\frac{(\text x)^2+1}{2.\text x })^2-(\frac{(\text x )^2-1}{2.\text x })^2 \\\\\\ \text E = \frac{\text x^4+2\text x^2+1 }{4\text x^2}-\frac{(\text x^4-2\text x^2+1)}{4\text x^2} \\\\\\ \text E = \frac{\text x^4+2\text x^2+1-\text x^4+2\text x^2-1}{4\text x^2} \\\\\\ \text E = \frac{4\text x^2}{4\text x^2}$

Portanto :

$\huge\boxed{\text E = 1\ }\checkmark$