Question

giúp em bài 6 với ạ, em cảm ơn

Answer 1

Bài 6:

Ta có:

\(sin^2x+cos^2x=1\)

\(\Leftrightarrow cos^2x=1-sin^2x\)

\(\Leftrightarrow cos^2x=1-\left(\dfrac{1}{3}\right)^2=\dfrac{8}{9}\)

\(\Leftrightarrow cosx=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}\) 

Mà: \(tanx=\dfrac{sinx}{cosx}\)

\(\Leftrightarrow tanx=\dfrac{1}{3}:\dfrac{2\sqrt{2}}{3}=\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{4}\) 

\(\Leftrightarrow\text{c}otx=\dfrac{1}{tanx}=1:\dfrac{\sqrt{2}}{4}=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)