Question

Dada a função f(x) = 1+ senx/cosx + cosx/1+senx, calcular f(π/6)

Answer 1

Como faltam parênteses, vou supor que a função é a seguinte:

$f(x)=\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}$

Como π/6 rad ≡ 30°, o seno e o cosseno de π/6 rad são iguais aos de 30°, então:

$f\left(\frac{\pi}{6}\right)=\dfrac{1+\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)}+\dfrac{\cos\left(\frac{\pi}{6}\right)}{1+\sin\left(\frac{\pi}{6}\right)}$

$f\left(\frac{\pi}{6}\right)=\dfrac{1+\left(\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}+\dfrac{\frac{\sqrt{3}}{2}}{1+\left(\frac{1}{2}\right)}$

$f\left(\frac{\pi}{6}\right)=\dfrac{\frac{3}{2}}{\frac{\sqrt{3}}{2}}+\dfrac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}$

$f\left(\frac{\pi}{6}\right)=\dfrac{3}{2}\cdot\dfrac{2}{\sqrt{3}}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{2}{3}$

$f\left(\frac{\pi}{6}\right)=\dfrac{3}{\sqrt{3}}+\dfrac{\sqrt{3}}{3}$

$f\left(\frac{\pi}{6}\right)=\dfrac{3\sqrt{3}}{3}+\dfrac{\sqrt{3}}{3}$

$f\left(\frac{\pi}{6}\right)=\dfrac{4\sqrt{3}}{3}$

E aí está!