Question

0 < X < pi/2  e  cossec x = 3, obtenha tg x  e cos x

Answer 1

Olá, Arthur.

 

$\text{<var>cossec }x=\frac1{\text{sen } x}=3 \Rightarrow \text{sen }x = \frac13 \\\\ \text{sen }^2 x+\cos^2 x=1 \Rightarrow \frac19+\cos^2x=1 \Rightarrow\cos x=\sqrt{1-\frac19}=\sqrt{\frac89} \Rightarrow \\\\ \cos x=\sqrt\frac{2^3}{3^2} \Rightarrow \boxed{\cos x=\frac{2\sqrt2}3} \\\\\\ \text{tg }x=\frac{\text{sen } x}{\cos x}=\frac13 \cdot \frac3{2\sqrt2}=\frac{\sqrt2}{2\sqrt2\sqrt2} \Rightarrow \boxed{\text{tg }x=\frac{\sqrt2}4}</var>$